80 results found for "Quadratic Equations" in Class 10.
Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(3x^2+10x+3=0\) के मूलों की प्रकृति क्या है?
What is the nature of roots of \(3x^2+10x+3=0\)?
#quadratic equations
#nature of roots
#discriminant
#distinct roots
A दो वास्तविक और असमान / two real and distinct
B दो वास्तविक और समान / two real and equal
C वास्तविक मूल नहीं / no real roots
D दोनों मूल (0) / both roots are (0)
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और असमान / two real and distinct
Step 1
Concept
(D=102 -4(3)(3)=64>0). Therefore the roots are real and distinct.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और असमान / two real and distinct. (D=102 -4(3)(3)=64>0). Therefore the roots are real and distinct.
Step 3
Exam Tip
(D=102 -4(3)(3)=64>0) है। इसलिए मूल वास्तविक और असमान हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
कथन पढ़िए: \(x^2+1=0\) के वास्तविक मूल नहीं हैं क्योंकि (D<0)। यह कथन कैसा है?
Read the statement: \(x^2+1=0\) has no real roots because (D<0). How is this statement?
#quadratic equations
#reasoning
#no real roots
#assertion
A सही / correct
B गलट / incorrect
C तभी सही जब (x=1) / correct only when (x=1)
D तभी सही जब (D=0) / correct only when (D=0)
Explanation opens after your attempt
Correct Answer
A. सही / correct
Step 1
Concept
(D=02 -4(1)(1)=-4<0). Hence the statement is correct and there are no real roots.
Step 2
Why this answer is correct
The correct answer is A. सही / correct. (D=02 -4(1)(1)=-4<0). Hence the statement is correct and there are no real roots.
Step 3
Exam Tip
(D=02 -4(1)(1)=-4<0) है। इसलिए कथन सही है और वास्तविक मूल नहीं हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(5x^2-4x+t=0\) के वास्तविक और समान मूलों के लिए (t) क्या होगा?
What is (t) for real and equal roots of \(5x^2-4x+t=0\)?
#quadratic equations
#parameter
#equal roots
#fraction
A \(\frac{4}{5}\)
B \(\frac{5}{4}\)
C (4)
D (5)
Explanation opens after your attempt
Correct Answer
A. \(\frac{4}{5}\)
Step 1
Concept
(D=(-4)2 -4(5)t=0) gives \(t=\frac{4}{5}\). Always set (D=0) for equal roots.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4}{5}\). (D=(-4)2 -4(5)t=0) gives \(t=\frac{4}{5}\). Always set (D=0) for equal roots.
Step 3
Exam Tip
(D=(-4)2 -4(5)t=0) से \(t=\frac{4}{5}\) मिलता है। समान मूलों के लिए हमेशा (D=0) रखें।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2-3x+s=0\) के समान वास्तविक मूलों के लिए (s) का मान क्या होगा?
What is the value of (s) for equal real roots of \(x^2-3x+s=0\)?
#quadratic equations
#parameter
#equal roots
#fraction
A \(\frac{9}{4}\)
B \(\frac{3}{4}\)
C (9)
D (3)
Explanation opens after your attempt
Correct Answer
A. \(\frac{9}{4}\)
Step 1
Concept
For equal roots (D=0), so ((-3)2 -4s=0) gives \(s=\frac{9}{4}\). Do not fear fractional answers.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9}{4}\). For equal roots (D=0), so ((-3)2 -4s=0) gives \(s=\frac{9}{4}\). Do not fear fractional answers.
Step 3
Exam Tip
समान मूलों के लिए (D=0), इसलिए ((-3)2 -4s=0) से \(s=\frac{9}{4}\)। भिन्न उत्तर से न डरें।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2-4x+8=0\) के मूलों की प्रकृति बताइए।
State the nature of roots of \(x^2-4x+8=0\).
#quadratic equations
#numerical
#no real roots
A वास्तविक मूल नहीं / no real roots
B दो वास्तविक और असमान / two real and distinct
C दो वास्तविक और समान / two real and equal
D दोनों मूल (8) / both roots are (8)
Explanation opens after your attempt
Correct Answer
A. वास्तविक मूल नहीं / no real roots
Step 1
Concept
(D=(-4)2 -4(1)(8)=-16<0). So there are no real roots.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं / no real roots. (D=(-4)2 -4(1)(8)=-16<0). So there are no real roots.
Step 3
Exam Tip
(D=(-4)2 -4(1)(8)=-16<0) है। इसलिए कोई वास्तविक मूल नहीं है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2+14x+49=0\) के लिए सही उत्तर चुनिए।
Choose the correct answer for \(x^2+14x+49=0\).
#quadratic equations
#numerical
#equal roots
A दो वास्तविक और समान / two real and equal
B दो वास्तविक और असमान / two real and distinct
C वास्तविक मूल नहीं / no real roots
D द्विघात नहीं / not quadratic
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और समान / two real and equal
Step 1
Concept
(D=142 -4(1)(49)=0). Hence the roots are equal and real.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और समान / two real and equal. (D=142 -4(1)(49)=0). Hence the roots are equal and real.
Step 3
Exam Tip
(D=142 -4(1)(49)=0) है। अतः मूल समान वास्तविक हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2-11x+30=0\) के लिए मूलों की प्रकृति क्या होगी?
What will be the nature of roots for \(x^2-11x+30=0\)?
#quadratic equations
#numerical
#distinct roots
A दो वास्तविक और असमान / two real and distinct
B दो वास्तविक और समान / two real and equal
C वास्तविक मूल नहीं / no real roots
D एक मूल शून्य / one root is zero
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और असमान / two real and distinct
Step 1
Concept
(D=(-11)2 -4(1)(30)=1>0). Therefore the roots will be real and distinct.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और असमान / two real and distinct. (D=(-11)2 -4(1)(30)=1>0). Therefore the roots will be real and distinct.
Step 3
Exam Tip
(D=(-11)2 -4(1)(30)=1>0) है। इसलिए मूल वास्तविक और असमान होंगे।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(2x^2+4x+2=0\) के लिए विविक्तकर का चिन्ह क्या है?
What is the sign of the discriminant for \(2x^2+4x+2=0\)?
#quadratic equations
#sign of discriminant
#equal roots
A शून्य / zero
B धनात्मक / positive
C ऋणात्मक / negative
D अज्ञात / unknown
Explanation opens after your attempt
Correct Answer
A. शून्य / zero
Step 1
Concept
(D=42 -4(2)(2)=0). So this is the case of equal real roots.
Step 2
Why this answer is correct
The correct answer is A. शून्य / zero. (D=42 -4(2)(2)=0). So this is the case of equal real roots.
Step 3
Exam Tip
(D=42 -4(2)(2)=0) है। इसलिए यह समान वास्तविक मूलों की स्थिति है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(3x^2-2x+4=0\) के लिए विविक्तकर का चिन्ह क्या है?
What is the sign of the discriminant for \(3x^2-2x+4=0\)?
#quadratic equations
#sign of discriminant
#no real roots
A ऋणात्मक / negative
B धनात्मक / positive
C शून्य / zero
D निश्चित नहीं / not fixed
Explanation opens after your attempt
Correct Answer
A. ऋणात्मक / negative
Step 1
Concept
(D=(-2)2 -4(3)(4)=-44<0). Hence the discriminant is negative.
Step 2
Why this answer is correct
The correct answer is A. ऋणात्मक / negative. (D=(-2)2 -4(3)(4)=-44<0). Hence the discriminant is negative.
Step 3
Exam Tip
(D=(-2)2 -4(3)(4)=-44<0) है। अतः विविक्तकर ऋणात्मक है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2+5x+6=0\) के लिए विविक्तकर और प्रकृति का सही युग्म कौन सा है?
Which pair of discriminant and nature is correct for \(x^2+5x+6=0\)?
#quadratic equations
#discriminant nature
#pair
A (D=1), दो वास्तविक और असमान / (D=1), two real and distinct
B (D=0), दो वास्तविक और समान / (D=0), two real and equal
C (D=-1), वास्तविक मूल नहीं / (D=-1), no real roots
D (D=25), दो वास्तविक और समान / (D=25), two real and equal
Explanation opens after your attempt
Correct Answer
A. (D=1), दो वास्तविक और असमान / (D=1), two real and distinct
Step 1
Concept
(D=52 -4(1)(6)=1). A positive (D) gives real and distinct roots.
Step 2
Why this answer is correct
The correct answer is A. (D=1), दो वास्तविक और असमान / (D=1), two real and distinct. (D=52 -4(1)(6)=1). A positive (D) gives real and distinct roots.
Step 3
Exam Tip
(D=52 -4(1)(6)=1) है। धनात्मक (D) से वास्तविक और असमान मूल होते हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(4x^2+4x+5=0\) के लिए सही प्रकृति कौन सी है?
Which is the correct nature for \(4x^2+4x+5=0\)?
#quadratic equations
#numerical
#no real roots
A वास्तविक मूल नहीं / no real roots
B दो वास्तविक और समान / two real and equal
C दो वास्तविक और असमान / two real and distinct
D दोनों मूल (5) / both roots are (5)
Explanation opens after your attempt
Correct Answer
A. वास्तविक मूल नहीं / no real roots
Step 1
Concept
(D=42 -4(4)(5)=-64<0). Therefore it has no real roots.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं / no real roots. (D=42 -4(4)(5)=-64<0). Therefore it has no real roots.
Step 3
Exam Tip
(D=42 -4(4)(5)=-64<0) है। इसलिए इसके वास्तविक मूल नहीं हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(2x^2-7x+3=0\) के मूलों की प्रकृति क्या है?
What is the nature of roots of \(2x^2-7x+3=0\)?
#quadratic equations
#numerical
#distinct roots
A दो वास्तविक और असमान / two real and distinct
B दो वास्तविक और समान / two real and equal
C वास्तविक मूल नहीं / no real roots
D दोनों मूल ऋणात्मक और समान / both roots negative and equal
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और असमान / two real and distinct
Step 1
Concept
(D=(-7)2 -4(2)(3)=25>0). Hence the roots are real and distinct.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और असमान / two real and distinct. (D=(-7)2 -4(2)(3)=25>0). Hence the roots are real and distinct.
Step 3
Exam Tip
(D=(-7)2 -4(2)(3)=25>0) है। इसलिए मूल वास्तविक और असमान हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2-6x+9=0\) के लिए कौन सा कथन सही है?
Which statement is correct for \(x^2-6x+9=0\)?
#quadratic equations
#numerical
#equal roots
A दो समान वास्तविक मूल हैं / it has two equal real roots
B दो असमान वास्तविक मूल हैं / it has two distinct real roots
C कोई वास्तविक मूल नहीं है / it has no real roots
D विविक्तकर (9) है / the discriminant is (9)
Explanation opens after your attempt
Correct Answer
A. दो समान वास्तविक मूल हैं / it has two equal real roots
Step 1
Concept
(D=(-6)2 -4(1)(9)=0). Therefore it has two equal real roots.
Step 2
Why this answer is correct
The correct answer is A. दो समान वास्तविक मूल हैं / it has two equal real roots. (D=(-6)2 -4(1)(9)=0). Therefore it has two equal real roots.
Step 3
Exam Tip
(D=(-6)2 -4(1)(9)=0) है। इसलिए दो समान वास्तविक मूल हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
यदि किसी द्विघात समीकरण के वास्तविक मूल नहीं हैं, तो (D) के लिए सही शर्त क्या है?
If a quadratic equation has no real roots, what is the correct condition for (D)?
#quadratic equations
#concept
#no real roots
A (D<0)
B (D=0)
C (D>0)
D \(D\geq0\)
Explanation opens after your attempt
Step 1
Concept
When there are no real roots, (D<0). The condition \(D\geq0\) indicates real roots.
Step 2
Why this answer is correct
The correct answer is A. (D<0). When there are no real roots, (D<0). The condition \(D\geq0\) indicates real roots.
Step 3
Exam Tip
वास्तविक मूल न होने पर (D<0) होता है। \(D\geq0\) वास्तविक मूलों की स्थिति बताता है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
यदि किसी द्विघात समीकरण के दो असमान वास्तविक मूल हैं, तो (D) कैसा होगा?
If a quadratic equation has two distinct real roots, how will (D) be?
#quadratic equations
#concept
#distinct roots
A (D>0)
B (D=0)
C (D<0)
D \(D\leq0\)
Explanation opens after your attempt
Step 1
Concept
For distinct real roots, (D>0). Do not add the equality sign by mistake.
Step 2
Why this answer is correct
The correct answer is A. (D>0). For distinct real roots, (D>0). Do not add the equality sign by mistake.
Step 3
Exam Tip
असमान वास्तविक मूलों के लिए (D>0) होता है। बराबर का चिन्ह गलती से न लगाएं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
यदि किसी द्विघात समीकरण के दो समान वास्तविक मूल हैं, तो (D) का मान क्या होगा?
If a quadratic equation has two equal real roots, what will be the value of (D)?
#quadratic equations
#concept
#equal roots
A (0)
B (1)
C (-1)
D (4)
Explanation opens after your attempt
Step 1
Concept
For equal real roots, (D=0) is necessary. This is the main rule for nature questions.
Step 2
Why this answer is correct
The correct answer is A. (0). For equal real roots, (D=0) is necessary. This is the main rule for nature questions.
Step 3
Exam Tip
समान वास्तविक मूलों के लिए (D=0) अनिवार्य है। प्रकृति वाले प्रश्नों में यही मुख्य नियम है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(2x^2+8=0\) के मूलों की प्रकृति क्या होगी?
What will be the nature of roots of \(2x^2+8=0\)?
#quadratic equations
#zero b
#no real roots
A वास्तविक मूल नहीं / no real roots
B दो वास्तविक और असमान / two real and distinct
C दो वास्तविक और समान / two real and equal
D दोनों मूल (-2) / both roots are (-2)
Explanation opens after your attempt
Correct Answer
A. वास्तविक मूल नहीं / no real roots
Step 1
Concept
(D=02 -4(2)(8)=-64<0). Therefore the equation will have no real roots.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं / no real roots. (D=02 -4(2)(8)=-64<0). Therefore the equation will have no real roots.
Step 3
Exam Tip
(D=02 -4(2)(8)=-64<0) है। इसलिए समीकरण के वास्तविक मूल नहीं होंगे।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(4x^2-1=0\) के मूल कैसे होंगे?
How will the roots of \(4x^2-1=0\) be?
#quadratic equations
#zero b
#distinct roots
A दो वास्तविक और असमान / two real and distinct
B दो वास्तविक और समान / two real and equal
C वास्तविक मूल नहीं / no real roots
D दोनों मूल (1) / both roots are (1)
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और असमान / two real and distinct
Step 1
Concept
(D=02 -4(4)(-1)=16>0). Hence two distinct real roots will be obtained.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और असमान / two real and distinct. (D=02 -4(4)(-1)=16>0). Hence two distinct real roots will be obtained.
Step 3
Exam Tip
(D=02 -4(4)(-1)=16>0) है। इसलिए दो असमान वास्तविक मूल मिलेंगे।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2+9=0\) के लिए सही विकल्प चुनिए।
Choose the correct option for \(x^2+9=0\).
#quadratic equations
#no real roots
#zero b
A वास्तविक मूल नहीं / no real roots
B दो वास्तविक और असमान / two real and distinct
C दो वास्तविक और समान / two real and equal
D एक मूल (9) है / one root is (9)
Explanation opens after your attempt
Correct Answer
A. वास्तविक मूल नहीं / no real roots
Step 1
Concept
Here (D=02 -4(1)(9)=-36<0). Therefore there are no real roots.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं / no real roots. Here (D=02 -4(1)(9)=-36<0). Therefore there are no real roots.
Step 3
Exam Tip
यहां (D=02 -4(1)(9)=-36<0) है। इसलिए वास्तविक मूल नहीं हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2-9=0\) के मूलों की प्रकृति क्या है?
What is the nature of roots of \(x^2-9=0\)?
#quadratic equations
#numerical
#distinct roots
A दो वास्तविक और असमान / two real and distinct
B दो वास्तविक और समान / two real and equal
C वास्तविक मूल नहीं / no real roots
D दोनों मूल (0) / both roots are (0)
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और असमान / two real and distinct
Step 1
Concept
Here (D=02 -4(1)(-9)=36>0). So the roots are real and distinct.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और असमान / two real and distinct. Here (D=02 -4(1)(-9)=36>0). So the roots are real and distinct.
Step 3
Exam Tip
यहां (D=02 -4(1)(-9)=36>0) है। इसलिए मूल वास्तविक और असमान हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2+12x+36=0\) के बारे में सही कथन कौन सा है?
Which statement is correct about \(x^2+12x+36=0\)?
#quadratic equations
#perfect square
#equal roots
A यह समान वास्तविक मूल रखता है / it has equal real roots
B यह कोई वास्तविक मूल नहीं रखता / it has no real roots
C यह असमान वास्तविक मूल रखता है / it has distinct real roots
D यह द्विघात समीकरण नहीं है / it is not a quadratic equation
Explanation opens after your attempt
Correct Answer
A. यह समान वास्तविक मूल रखता है / it has equal real roots
Step 1
Concept
(x-2 +12x+36=(x+6)2 ). Being a perfect square, it has equal real roots.
Step 2
Why this answer is correct
The correct answer is A. यह समान वास्तविक मूल रखता है / it has equal real roots. (x-2 +12x+36=(x+6)2 ). Being a perfect square, it has equal real roots.
Step 3
Exam Tip
(x-2 +12x+36=(x+6)2 ) है। पूर्ण वर्ग होने से मूल समान वास्तविक हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(2x^2+3x+\lambda=0\) के वास्तविक और असमान मूलों के लिए कौन सी शर्त सही है?
For \(2x^2+3x+\lambda=0\) to have real and distinct roots, which condition is correct?
#quadratic equations
#parameter
#distinct roots
#fraction
A \(\lambda<\frac{9}{8}\)
B \(\lambda=\frac{9}{8}\)
C \(\lambda>\frac{9}{8}\)
D \(\lambda=\frac{8}{9}\)
Explanation opens after your attempt
Correct Answer
A. \(\lambda<\frac{9}{8}\)
Step 1
Concept
(D=32 -4(2)\lambda=9-8\lambda). From (D>0), we get \(\lambda<\frac{9}{8}\).
Step 2
Why this answer is correct
The correct answer is A. \(\lambda<\frac{9}{8}\). (D=32 -4(2)\lambda=9-8\lambda). From (D>0), we get \(\lambda<\frac{9}{8}\).
Step 3
Exam Tip
(D=32 -4(2)\lambda=9-8\lambda) है। (D>0) से \(\lambda<\frac{9}{8}\) मिलता है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(3x^2+kx+12=0\) में समान वास्तविक मूलों के लिए (k) का कौन सा मान सही हो सकता है?
Which value of (k) can give equal real roots in \(3x^2+kx+12=0\)?
#quadratic equations
#parameter
#equal roots
A (12)
B (6)
C (9)
D (3)
Explanation opens after your attempt
Step 1
Concept
For equal roots (k-2 -4(3)(12)=0), so (k=12) or (k=-12). Among the options, (12) is correct.
Step 2
Why this answer is correct
The correct answer is A. (12). For equal roots (k-2 -4(3)(12)=0), so (k=12) or (k=-12). Among the options, (12) is correct.
Step 3
Exam Tip
समान मूलों के लिए (k-2 -4(3)(12)=0), इसलिए (k=12) या (k=-12)। दिए विकल्पों में (12) सही है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2-10x+r=0\) में समान वास्तविक मूलों के लिए (r) का मान चुनिए।
Choose the value of (r) for equal real roots in \(x^2-10x+r=0\).
#quadratic equations
#parameter
#equal roots
A (25)
B (10)
C (20)
D (100)
Explanation opens after your attempt
Step 1
Concept
(D=(-10)2 -4r=0) gives (r=25). Equal roots have zero discriminant.
Step 2
Why this answer is correct
The correct answer is A. (25). (D=(-10)2 -4r=0) gives (r=25). Equal roots have zero discriminant.
Step 3
Exam Tip
(D=(-10)2 -4r=0) से (r=25) मिलता है। समान मूलों में विविक्तकर शून्य होता है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2+4x+p=0\) के वास्तविक मूल न होने के लिए कौन सी शर्त सही है?
For \(x^2+4x+p=0\) to have no real roots, which condition is correct?
#quadratic equations
#parameter
#no real roots
#inequality
A (p>4)
B (p=4)
C (p<4)
D (p=0)
Explanation opens after your attempt
Step 1
Concept
For no real roots (D<0), so (16-4p<0) gives (p>4). A negative discriminant gives no real roots.
Step 2
Why this answer is correct
The correct answer is A. (p>4). For no real roots (D<0), so (16-4p<0) gives (p>4). A negative discriminant gives no real roots.
Step 3
Exam Tip
वास्तविक मूल न होने के लिए (D<0), इसलिए (16-4p<0) से (p>4)। ऋणात्मक विविक्तकर पर वास्तविक मूल नहीं होते।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2-2x+n=0\) के दो वास्तविक और असमान मूल होने के लिए कौन सी शर्त सही है?
For \(x^2-2x+n=0\) to have two real and distinct roots, which condition is correct?
#quadratic equations
#parameter
#distinct roots
#inequality
A (n<1)
B (n=1)
C (n>1)
D (n=2)
Explanation opens after your attempt
Step 1
Concept
For distinct real roots (D>0), so ((-2)2 -4n>0) gives (n<1). Use a strict inequality for distinct roots.
Step 2
Why this answer is correct
The correct answer is A. (n<1). For distinct real roots (D>0), so ((-2)2 -4n>0) gives (n<1). Use a strict inequality for distinct roots.
Step 3
Exam Tip
असमान वास्तविक मूलों के लिए (D>0), इसलिए ((-2)2 -4n>0) से (n<1)। असमान के लिए कड़ाई वाली असमता लगती है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(7x^2+2x+3=0\) के लिए सही निष्कर्ष क्या है?
What is the correct conclusion for \(7x^2+2x+3=0\)?
#quadratic equations
#no real roots
#numerical
A वास्तविक मूल नहीं / no real roots
B दो वास्तविक और असमान / two real and distinct
C दो वास्तविक और समान / two real and equal
D एक मूल (-3) है / one root is (-3)
Explanation opens after your attempt
Correct Answer
A. वास्तविक मूल नहीं / no real roots
Step 1
Concept
(D=22 -4(7)(3)=-80<0). So this equation has no real roots.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं / no real roots. (D=22 -4(7)(3)=-80<0). So this equation has no real roots.
Step 3
Exam Tip
(D=22 -4(7)(3)=-80<0) है। इसलिए यह समीकरण वास्तविक मूल नहीं रखता।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(6x^2-5x+1=0\) का विविक्तकर क्या है?
What is the discriminant of \(6x^2-5x+1=0\)?
#quadratic equations
#discriminant value
#numerical
A (1)
B (49)
C (-1)
D (0)
Explanation opens after your attempt
Step 1
Concept
(D=(-5)2 -4(6)(1)=1). This gives two real and distinct roots.
Step 2
Why this answer is correct
The correct answer is A. (1). (D=(-5)2 -4(6)(1)=1). This gives two real and distinct roots.
Step 3
Exam Tip
(D=(-5)2 -4(6)(1)=1) है। इससे दो वास्तविक और असमान मूल होंगे।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2+x+3=0\) के मूलों की प्रकृति क्या होगी?
What will be the nature of roots of \(x^2+x+3=0\)?
#quadratic equations
#numerical
#no real roots
A वास्तविक मूल नहीं / no real roots
B दो वास्तविक और असमान / two real and distinct
C दो वास्तविक और समान / two real and equal
D एक मूल (1) होगा / one root will be (1)
Explanation opens after your attempt
Correct Answer
A. वास्तविक मूल नहीं / no real roots
Step 1
Concept
(D=12 -4(1)(3)=-11<0). Hence real roots will not be obtained.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं / no real roots. (D=12 -4(1)(3)=-11<0). Hence real roots will not be obtained.
Step 3
Exam Tip
(D=12 -4(1)(3)=-11<0) है। इसलिए वास्तविक मूल नहीं मिलेंगे।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(9x^2-6x+1=0\) में मूलों की प्रकृति क्या है?
What is the nature of roots in \(9x^2-6x+1=0\)?
#quadratic equations
#numerical
#equal roots
A दो वास्तविक और समान / two real and equal
B दो वास्तविक और असमान / two real and distinct
C वास्तविक मूल नहीं / no real roots
D दो काल्पनिक और समान / two imaginary and equal
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और समान / two real and equal
Step 1
Concept
(D=(-6)2 -4(9)(1)=0). Therefore this is a case of equal real roots.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और समान / two real and equal. (D=(-6)2 -4(9)(1)=0). Therefore this is a case of equal real roots.
Step 3
Exam Tip
(D=(-6)2 -4(9)(1)=0) है। इसलिए यह समान वास्तविक मूलों का मामला है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(2x^2+5x+2=0\) के लिए सही प्रकृति चुनिए।
Choose the correct nature for \(2x^2+5x+2=0\).
#quadratic equations
#numerical
#distinct roots
A दो वास्तविक और असमान / two real and distinct
B दो वास्तविक और समान / two real and equal
C वास्तविक मूल नहीं / no real roots
D दोनों मूल शून्य / both roots are zero
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और असमान / two real and distinct
Step 1
Concept
(D=52 -4(2)(2)=9>0). Hence the roots are real and distinct.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और असमान / two real and distinct. (D=52 -4(2)(2)=9>0). Hence the roots are real and distinct.
Step 3
Exam Tip
(D=52 -4(2)(2)=9>0) है। इसलिए मूल वास्तविक और असमान हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
यदि (D=0) और \(a\neq0\) हो तो द्विघात समीकरण में कितने अलग-अलग वास्तविक मूल होंगे?
If (D=0) and \(a\neq0\), how many distinct real roots will the quadratic equation have?
#quadratic equations
#equal roots
#distinct count
A (1)
B (2)
C (0)
D (3)
Explanation opens after your attempt
Step 1
Concept
At (D=0), both roots are equal, so the number of distinct real roots is (1). Remember the root is repeated.
Step 2
Why this answer is correct
The correct answer is A. (1). At (D=0), both roots are equal, so the number of distinct real roots is (1). Remember the root is repeated.
Step 3
Exam Tip
(D=0) पर दोनों मूल समान होते हैं, इसलिए अलग-अलग वास्तविक मूलों की संख्या (1) है। ध्यान रखें मूल दो बार दोहरता है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
यदि (D=-7) हो तो मूलों की प्रकृति क्या होगी?
If (D=-7), what will be the nature of roots?
#quadratic equations
#discriminant sign
#no real roots
A वास्तविक मूल नहीं / no real roots
B दो वास्तविक और असमान / two real and distinct
C दो वास्तविक और समान / two real and equal
D एक मूल (7) होगा / one root will be (7)
Explanation opens after your attempt
Correct Answer
A. वास्तविक मूल नहीं / no real roots
Step 1
Concept
(-7<0), so there will be no real roots. Never treat negative (D) as equal roots.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं / no real roots. (-7<0), so there will be no real roots. Never treat negative (D) as equal roots.
Step 3
Exam Tip
(-7<0) है, इसलिए वास्तविक मूल नहीं होंगे। ऋणात्मक (D) को कभी समान मूल न मानें।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
यदि (D=25) हो तो द्विघात समीकरण के मूलों के बारे में सही बात क्या है?
If (D=25), which statement about the roots of a quadratic equation is correct?
#quadratic equations
#discriminant sign
#distinct roots
A मूल वास्तविक और असमान होंगे / roots will be real and distinct
B मूल वास्तविक और समान होंगे / roots will be real and equal
C वास्तविक मूल नहीं होंगे / roots will not be real
D मूल हमेशा शून्य होंगे / roots will always be zero
Explanation opens after your attempt
Correct Answer
A. मूल वास्तविक और असमान होंगे / roots will be real and distinct
Step 1
Concept
(25>0), so the roots will be real and distinct. The sign of (D) is enough for nature.
Step 2
Why this answer is correct
The correct answer is A. मूल वास्तविक और असमान होंगे / roots will be real and distinct. (25>0), so the roots will be real and distinct. The sign of (D) is enough for nature.
Step 3
Exam Tip
(25>0) है, इसलिए मूल वास्तविक और असमान होंगे। (D) का केवल चिन्ह प्रकृति बताने के लिए काफी है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(4x^2-12x+9=0\) को देखकर मूलों की प्रकृति क्या होगी?
By observing \(4x^2-12x+9=0\), what will be the nature of roots?
#quadratic equations
#perfect square
#equal roots
A दो वास्तविक और समान / two real and equal
B दो वास्तविक और असमान / two real and distinct
C वास्तविक मूल नहीं / no real roots
D मूल अपरिमेय और असमान / irrational and distinct roots
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और समान / two real and equal
Step 1
Concept
\(4x^2-12x+9\) is a perfect square ((2x-3)2 ). So the roots are equal and real.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और समान / two real and equal. \(4x^2-12x+9\) is a perfect square ((2x-3)2 ). So the roots are equal and real.
Step 3
Exam Tip
\(4x^2-12x+9\) एक पूर्ण वर्ग ((2x-3)2 ) है। इसलिए मूल समान वास्तविक हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
किस समीकरण के वास्तविक मूल नहीं होंगे?
Which equation will have no real roots?
#quadratic equations
#choose equation
#no real roots
A \(x^2+3x+5=0\)
B \(x^2-3x+2=0\)
C \(x^2-2x+1=0\)
D \(2x^2-5x+2=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+3x+5=0\)
Step 1
Concept
For the first equation, (D=32 -4(1)(5)=-11<0). A negative discriminant gives no real roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3x+5=0\). For the first equation, (D=32 -4(1)(5)=-11<0). A negative discriminant gives no real roots.
Step 3
Exam Tip
पहले समीकरण में (D=32 -4(1)(5)=-11<0) है। ऋणात्मक विविक्तकर वास्तविक मूल नहीं देता।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
किस समीकरण के दो वास्तविक और समान मूल होंगे?
Which equation will have two real and equal roots?
#quadratic equations
#choose equation
#equal roots
A \(x^2+10x+25=0\)
B \(x^2+10x+24=0\)
C \(x^2+10x+30=0\)
D \(2x^2+10x+25=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+10x+25=0\)
Step 1
Concept
In the first equation, (D=102 -4(1)(25)=0). A perfect square form often gives equal roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+10x+25=0\). In the first equation, (D=102 -4(1)(25)=0). A perfect square form often gives equal roots.
Step 3
Exam Tip
पहले समीकरण में (D=102 -4(1)(25)=0) है। पूर्ण वर्ग रूप अक्सर समान मूल देता है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
किस समीकरण के दो वास्तविक और असमान मूल होंगे?
Which equation will have two real and distinct roots?
#quadratic equations
#choose equation
#distinct roots
A \(x^2-7x+10=0\)
B \(x^2+2x+1=0\)
C \(x^2+4x+8=0\)
D \(4x^2+4x+1=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-7x+10=0\)
Step 1
Concept
For the first equation, (D=(-7)2 -4(1)(10)=9>0). Hence its roots are real and distinct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-7x+10=0\). For the first equation, (D=(-7)2 -4(1)(10)=9>0). Hence its roots are real and distinct.
Step 3
Exam Tip
पहले समीकरण में (D=(-7)2 -4(1)(10)=9>0) है। इसलिए उसके मूल वास्तविक और असमान हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2+kx+16=0\) के समान वास्तविक मूल होने के लिए (k) का कौन सा मान सही हो सकता है?
Which value of (k) can make \(x^2+kx+16=0\) have equal real roots?
#quadratic equations
#parameter
#equal roots
A (8)
B (4)
C (16)
D (2)
Explanation opens after your attempt
Step 1
Concept
For equal roots \(k^2-64=0\), so (k=8) or (k=-8). Among the options, (8) is correct.
Step 2
Why this answer is correct
The correct answer is A. (8). For equal roots \(k^2-64=0\), so (k=8) or (k=-8). Among the options, (8) is correct.
Step 3
Exam Tip
समान मूलों के लिए \(k^2-64=0\) होता है, इसलिए (k=8) या (k=-8)। दिए विकल्पों में (8) सही है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2-8x+m=0\) में समान वास्तविक मूलों के लिए (m) क्या होगा?
In \(x^2-8x+m=0\), what is (m) for equal real roots?
#quadratic equations
#parameter
#equal roots
A (16)
B (8)
C (4)
D (64)
Explanation opens after your attempt
Step 1
Concept
(D=(-8)2 -4m=0) gives (m=16). Keep the discriminant zero for equal roots.
Step 2
Why this answer is correct
The correct answer is A. (16). (D=(-8)2 -4m=0) gives (m=16). Keep the discriminant zero for equal roots.
Step 3
Exam Tip
(D=(-8)2 -4m=0) से (m=16) मिलता है। बराबर मूलों के लिए विविक्तकर शून्य रखें।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2+6x+k=0\) के समान वास्तविक मूल होने के लिए (k) का मान क्या होगा?
For \(x^2+6x+k=0\) to have equal real roots, what should be the value of (k)?
#quadratic equations
#parameter
#equal roots
A (9)
B (6)
C (12)
D (36)
Explanation opens after your attempt
Step 1
Concept
For equal roots (D=0), so \(6^2-4k=0\) gives (k=9). Use (D=0) in such questions.
Step 2
Why this answer is correct
The correct answer is A. (9). For equal roots (D=0), so \(6^2-4k=0\) gives (k=9). Use (D=0) in such questions.
Step 3
Exam Tip
समान मूलों के लिए (D=0), इसलिए \(6^2-4k=0\) से (k=9)। ऐसे प्रश्न में (D=0) लगाएं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(5x^2+x+1=0\) के लिए मूलों की प्रकृति चुनिए।
Choose the nature of roots for \(5x^2+x+1=0\).
#quadratic equations
#no real roots
#numerical
A वास्तविक मूल नहीं / no real roots
B दो वास्तविक और असमान / two real and distinct
C दो वास्तविक और समान / two real and equal
D दोनों मूल धनात्मक / both roots positive
Explanation opens after your attempt
Correct Answer
A. वास्तविक मूल नहीं / no real roots
Step 1
Concept
(D=12 -4(5)(1)=-19<0). Therefore the equation has no real roots.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं / no real roots. (D=12 -4(5)(1)=-19<0). Therefore the equation has no real roots.
Step 3
Exam Tip
(D=12 -4(5)(1)=-19<0) है। इसलिए कोई वास्तविक मूल नहीं है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(3x^2+6x+3=0\) के मूल किस प्रकार के होंगे?
What type of roots will the equation \(3x^2+6x+3=0\) have?
#quadratic equations
#equal roots
#discriminant
A दो वास्तविक और समान / two real and equal
B दो वास्तविक और असमान / two real and distinct
C वास्तविक मूल नहीं / no real roots
D एक धनात्मक और एक ऋणात्मक मूल / one positive and one negative root
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और समान / two real and equal
Step 1
Concept
(D=62 -4(3)(3)=0). Hence both roots will be equal and real.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और समान / two real and equal. (D=62 -4(3)(3)=0). Hence both roots will be equal and real.
Step 3
Exam Tip
(D=62 -4(3)(3)=0) है। इसलिए दोनों मूल समान वास्तविक होंगे।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(2x^2-3x+1=0\) के लिए विविक्तकर का मान क्या है?
What is the value of the discriminant for \(2x^2-3x+1=0\)?
#quadratic equations
#discriminant value
#numerical
A (1)
B (-1)
C (0)
D (25)
Explanation opens after your attempt
Step 1
Concept
The value is (D=(-3)2 -4(2)(1)=1). A positive discriminant gives distinct real roots.
Step 2
Why this answer is correct
The correct answer is A. (1). The value is (D=(-3)2 -4(2)(1)=1). A positive discriminant gives distinct real roots.
Step 3
Exam Tip
(D=(-3)2 -4(2)(1)=1) है। धनात्मक विविक्तकर असमान वास्तविक मूल देता है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2+2x+5=0\) के मूलों के बारे में सही कथन क्या है?
Which statement is correct about the roots of \(x^2+2x+5=0\)?
#quadratic equations
#numerical
#no real roots
A वास्तविक मूल नहीं / no real roots
B दो वास्तविक और समान / two real and equal
C दो वास्तविक और असमान / two real and distinct
D दोनों मूल (0) हैं / both roots are (0)
Explanation opens after your attempt
Correct Answer
A. वास्तविक मूल नहीं / no real roots
Step 1
Concept
Here (D=22 -4(1)(5)=-16<0). So there are no real roots.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं / no real roots. Here (D=22 -4(1)(5)=-16<0). So there are no real roots.
Step 3
Exam Tip
यहां (D=22 -4(1)(5)=-16<0) है। इसलिए वास्तविक मूल नहीं हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2-4x+4=0\) के मूलों की प्रकृति पहचानिए।
Identify the nature of roots of the equation \(x^2-4x+4=0\).
#quadratic equations
#numerical
#equal roots
A दो वास्तविक और समान / two real and equal
B दो वास्तविक और असमान / two real and distinct
C वास्तविक मूल नहीं / no real roots
D दो ऋणात्मक असमान मूल / two negative distinct roots
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और समान / two real and equal
Step 1
Concept
Here (D=(-4)2 -4(1)(4)=0). (D=0) means equal real roots.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और समान / two real and equal. Here (D=(-4)2 -4(1)(4)=0). (D=0) means equal real roots.
Step 3
Exam Tip
यहां (D=(-4)2 -4(1)(4)=0) है। (D=0) समान वास्तविक मूल बताता है।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
समीकरण \(x^2-5x+6=0\) के मूलों की प्रकृति क्या है?
What is the nature of roots of the equation \(x^2-5x+6=0\)?
#quadratic equations
#numerical
#distinct real roots
A दो वास्तविक और असमान / two real and distinct
B दो वास्तविक और समान / two real and equal
C वास्तविक मूल नहीं / no real roots
D केवल एक वास्तविक मूल / only one real root
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और असमान / two real and distinct
Step 1
Concept
Here (D=(-5)2 -4(1)(6)=1>0). So the roots are real and distinct.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और असमान / two real and distinct. Here (D=(-5)2 -4(1)(6)=1>0). So the roots are real and distinct.
Step 3
Exam Tip
यहां (D=(-5)2 -4(1)(6)=1>0) है। इसलिए मूल वास्तविक और असमान हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
यदि द्विघात समीकरण के लिए (D=0) हो तो मूल कैसे होंगे?
If (D=0) for a quadratic equation, how will the roots be?
#quadratic equations
#D equals zero
#equal roots
A दो वास्तविक और समान / two real and equal
B दो वास्तविक और असमान / two real and distinct
C कोई वास्तविक मूल नहीं / no real roots
D हमेशा धनात्मक / always positive
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और समान / two real and equal
Step 1
Concept
When (D=0), both roots are equal. These are also called repeated real roots.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और समान / two real and equal. When (D=0), both roots are equal. These are also called repeated real roots.
Step 3
Exam Tip
(D=0) होने पर दोनों मूल बराबर होते हैं। इसे दोहराए गए वास्तविक मूल भी कहते हैं।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
यदि द्विघात समीकरण के लिए (D>0) हो तो मूलों की प्रकृति क्या होगी?
If (D>0) for a quadratic equation, what will be the nature of roots?
#quadratic equations
#D greater than zero
#real roots
A दो वास्तविक और असमान / two real and distinct
B दो वास्तविक और समान / two real and equal
C कोई वास्तविक मूल नहीं / no real roots
D एक ही मूल और दूसरा शून्य / one root and the other zero
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और असमान / two real and distinct
Step 1
Concept
When (D>0), two different real roots are obtained. Check the sign first in exams.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और असमान / two real and distinct. When (D>0), two different real roots are obtained. Check the sign first in exams.
Step 3
Exam Tip
(D>0) होने पर दो अलग-अलग वास्तविक मूल मिलते हैं। चिन्ह देखकर प्रकृति तुरंत लिखें।
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Question
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
द्विघात समीकरण \(ax^2+bx+c=0\) के मूलों की प्रकृति जानने के लिए कौन सा व्यंजक प्रयोग होता है?
Which expression is used to determine the nature of roots of the quadratic equation \(ax^2+bx+c=0\)?
#quadratic equations
#nature of roots
#discriminant
#easy
A \(b^2-4ac\)
B \(b^2+4ac\)
C \(a^2-4bc\)
D \(c^2-4ab\)
Explanation opens after your attempt
Correct Answer
A. \(b^2-4ac\)
Step 1
Concept
The discriminant is \(D=b^2-4ac\). In exams first identify (a,b,c) correctly.
Step 2
Why this answer is correct
The correct answer is A. \(b^2-4ac\). The discriminant is \(D=b^2-4ac\). In exams first identify (a,b,c) correctly.
Step 3
Exam Tip
विविक्तकर \(D=b^2-4ac\) होता है। परीक्षा में पहले (a,b,c) सही पहचानें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(x^2-22x+79=0\) के मूल द्विघात सूत्र से क्या होंगे?
What are the roots of \(x^2-22x+79=0\) by quadratic formula?
#quadratic
#quadratic-formula
#application
A \(x=11\pm\sqrt{42}\)
B \(x=-11\pm\sqrt{42}\)
C \(x=22\pm\sqrt{42}\)
D \(x=\frac{11\pm\sqrt{42}}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x=11\pm\sqrt{42}\)
Step 1
Concept
Here (D=(-22)2 -4(1)(79)=168), so \(x=\frac{22\pm2\sqrt{42}}{2}=11\pm\sqrt{42}\). In exams, simplify (D) correctly.
Step 2
Why this answer is correct
The correct answer is A. \(x=11\pm\sqrt{42}\). Here (D=(-22)2 -4(1)(79)=168), so \(x=\frac{22\pm2\sqrt{42}}{2}=11\pm\sqrt{42}\). In exams, simplify (D) correctly.
Step 3
Exam Tip
यहां (D=(-22)2 -4(1)(79)=168), इसलिए \(x=\frac{22\pm2\sqrt{42}}{2}=11\pm\sqrt{42}\) है। परीक्षा में (D) को सही सरल करें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि ((x-7)(x-15)=26), तो मानक द्विघात समीकरण क्या होगा?
If ((x-7)(x-15)=26), what is the standard quadratic equation?
#quadratic
#standard-form
#application
A \(x^2-22x+79=0\)
B \(x^2-22x+131=0\)
C \(x^2+22x+79=0\)
D \(x^2-8x+105=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-22x+79=0\)
Step 1
Concept
((x-7)(x-15)=x-2 -22x+105), so \(x^2-22x+105=26\) gives \(x^2-22x+79=0\). In exams, bring all terms to one side after expansion.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-22x+79=0\). ((x-7)(x-15)=x-2 -22x+105), so \(x^2-22x+105=26\) gives \(x^2-22x+79=0\). In exams, bring all terms to one side after expansion.
Step 3
Exam Tip
((x-7)(x-15)=x-2 -22x+105), इसलिए \(x^2-22x+105=26\) से \(x^2-22x+79=0\) मिलता है। परीक्षा में विस्तार के बाद सभी पद एक तरफ लाएं।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
(x-2 -2(k+7)x+k-2 =0) के समान मूलों के लिए (k) क्या होगा?
What is (k) for equal roots of (x-2 -2(k+7)x+k-2 =0)?
#quadratic
#discriminant
#parameter
A \(k=-\frac{7}{2}\)
B \(k=\frac{7}{2}\)
C (k=-7)
D (k=7)
Explanation opens after your attempt
Correct Answer
A. \(k=-\frac{7}{2}\)
Step 1
Concept
(D=4(k+7)2 -4k-2 =0) gives ((k+7)2 =k-2 ), so (14k+49=0) and \(k=-\frac{7}{2}\). In exams, expand squares carefully.
Step 2
Why this answer is correct
The correct answer is A. \(k=-\frac{7}{2}\). (D=4(k+7)2 -4k-2 =0) gives ((k+7)2 =k-2 ), so (14k+49=0) and \(k=-\frac{7}{2}\). In exams, expand squares carefully.
Step 3
Exam Tip
(D=4(k+7)2 -4k-2 =0) से ((k+7)2 =k-2 ), इसलिए (14k+49=0) और \(k=-\frac{7}{2}\) है। परीक्षा में वर्ग फैलाते समय सावधानी रखें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(kx^2-26x+169=0\) के समान मूल हैं, तो (k) क्या होगा?
If \(kx^2-26x+169=0\) has equal roots, what is (k)?
#quadratic
#discriminant
#parameter
A (1)
B (2)
C (13)
D (169)
Explanation opens after your attempt
Step 1
Concept
For equal roots, (D=0), so (676-676k=0) and (k=1). In exams, keep (a=k) correctly.
Step 2
Why this answer is correct
The correct answer is A. (1). For equal roots, (D=0), so (676-676k=0) and (k=1). In exams, keep (a=k) correctly.
Step 3
Exam Tip
समान मूलों के लिए (D=0), इसलिए (676-676k=0) और (k=1) है। परीक्षा में (a=k) को ठीक से रखें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2-26x+165=0\) के मूल \(\alpha,\beta\) हैं, तो \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) क्या होगा?
If the roots of \(x^2-26x+165=0\) are \(\alpha,\beta\), what is \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\)?
#quadratic
#roots-expression
#ratio
A \( \frac{346}{165}\)
B \( \frac{676}{165}\)
C \( \frac{26}{165}\)
D \( \frac{165}{346}\)
Explanation opens after your attempt
Correct Answer
A. \( \frac{346}{165}\)
Step 1
Concept
\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=26\) and \(\alpha\beta=165\), so the value is \(\frac{676-330}{165}=\frac{346}{165}\). In exams, convert expressions into sum and product.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{346}{165}\). \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=26\) and \(\alpha\beta=165\), so the value is \(\frac{676-330}{165}=\frac{346}{165}\). In exams, convert expressions into sum and product.
Step 3
Exam Tip
\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), जहां \(\alpha+\beta=26\) और \(\alpha\beta=165\), इसलिए मान \(\frac{676-330}{165}=\frac{346}{165}\) है। परीक्षा में अभिव्यक्ति को योग और गुणनफल में बदलें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2-16x+48=0\) के मूल \(\alpha,\beta\) हैं, तो नए मूल \(\alpha+6,\beta+6\) वाला समीकरण कौनसा है?
If roots of \(x^2-16x+48=0\) are \(\alpha,\beta\), which equation has roots \(\alpha+6,\beta+6\)?
#quadratic
#transformed-roots
#equation
A \(x^2-28x+180=0\)
B \(x^2-16x+48=0\)
C \(x^2-4x+12=0\)
D \(x^2+28x+180=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-28x+180=0\)
Step 1
Concept
The roots are (4,12), so new roots are (10,18), and the equation is ((x-10)(x-18)=0). In exams, form the new roots and then the new equation.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-28x+180=0\). The roots are (4,12), so new roots are (10,18), and the equation is ((x-10)(x-18)=0). In exams, form the new roots and then the new equation.
Step 3
Exam Tip
मूल (4,12) हैं, इसलिए नए मूल (10,18) होंगे और समीकरण ((x-10)(x-18)=0) है। परीक्षा में नए मूल बनाकर नया समीकरण लिखें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(7x^2-14x+19=0\) में पूर्ण वर्ग रूप कौनसा सही है?
Which completed square form is correct for \(7x^2-14x+19=0\)?
#quadratic
#completing-square
#no-real-roots
A (7(x-1)2 +12=0)
B (7(x+1)2 +12=0)
C (7(x-1)2 -12=0)
D (7(x-2)2 +19=0)
Explanation opens after your attempt
Correct Answer
A. (7(x-1)2 +12=0)
Step 1
Concept
(7x-2 -14x+19=7(x-1)2 +12), so it cannot be zero for real (x). In exams, completed square form also shows the nature of roots.
Step 2
Why this answer is correct
The correct answer is A. (7(x-1)2 +12=0). (7x-2 -14x+19=7(x-1)2 +12), so it cannot be zero for real (x). In exams, completed square form also shows the nature of roots.
Step 3
Exam Tip
(7x-2 -14x+19=7(x-1)2 +12), इसलिए यह वास्तविक (x) के लिए शून्य नहीं हो सकता। परीक्षा में पूर्ण वर्ग रूप से भी मूलों की प्रकृति दिखती है।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(7x^2-14x+19=0\) के वास्तविक मूलों के बारे में सही कथन क्या है?
What is the correct statement about real roots of \(7x^2-14x+19=0\)?
#quadratic
#discriminant
#no-real-roots
A वास्तविक मूल नहीं हैं / There are no real roots
B दो समान वास्तविक मूल हैं / There are two equal real roots
C दो अलग वास्तविक मूल हैं / There are two distinct real roots
D एक मूल (0) है / One root is (0)
Explanation opens after your attempt
Correct Answer
A. वास्तविक मूल नहीं हैं / There are no real roots
Step 1
Concept
Here (D=(-14)2 -4(7)(19)=-336<0), so there are no real roots. In exams, (D<0) means no real roots.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं हैं / There are no real roots. Here (D=(-14)2 -4(7)(19)=-336<0), so there are no real roots. In exams, (D<0) means no real roots.
Step 3
Exam Tip
यहां (D=(-14)2 -4(7)(19)=-336<0), इसलिए वास्तविक मूल नहीं हैं। परीक्षा में (D<0) का अर्थ वास्तविक मूल नहीं है।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(x^2-13x+7=0\) के मूलों का अंतर क्या है?
What is the difference between the roots of \(x^2-13x+7=0\)?
#quadratic
#discriminant
#difference-of-roots
A \(\sqrt{141}\)
B (141)
C (13)
D (7)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{141}\)
Step 1
Concept
Here (D=(-13)2 -4(1)(7)=141), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{141}\). In exams, the difference of roots can be found directly from (D).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{141}\). Here (D=(-13)2 -4(1)(7)=141), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{141}\). In exams, the difference of roots can be found directly from (D).
Step 3
Exam Tip
यहां (D=(-13)2 -4(1)(7)=141), इसलिए मूलों का अंतर \(\frac{\sqrt{D}}{|a|}=\sqrt{141}\) है। परीक्षा में मूलों का अंतर सीधे (D) से मिल सकता है।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2+px+64=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?
If one root of \(x^2+px+64=0\) is double the other and both are negative, what is (p)?
#quadratic
#roots-relation
#parameter
A \(12\sqrt{2}\)
B \(-12\sqrt{2}\)
C \(16\sqrt{2}\)
D (24)
Explanation opens after your attempt
Correct Answer
A. \(12\sqrt{2}\)
Step 1
Concept
Let the roots be (-r) and (-2r), then \(2r^2=64\) gives \(r=4\sqrt{2}\), and \(p=3r=12\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 2
Why this answer is correct
The correct answer is A. \(12\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=64\) gives \(r=4\sqrt{2}\), and \(p=3r=12\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=64\) से \(r=4\sqrt{2}\) और \(p=3r=12\sqrt{2}\) है। परीक्षा में दोनों मूलों के चिन्ह ध्यान से रखें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2-12x-28=0\) के मूल \(\alpha,\beta\) हैं, तो \(\alpha^2\beta+\alpha\beta^2\) क्या होगा?
If the roots of \(x^2-12x-28=0\) are \(\alpha,\beta\), what is \(\alpha^2\beta+\alpha\beta^2\)?
#quadratic
#roots-expression
#advanced
A (-336)
B (336)
C (-28)
D (12)
Explanation opens after your attempt
Step 1
Concept
(\alpha-2 \beta+\alpha\beta-2 =\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=12\) and \(\alpha\beta=-28\), so the value is (-336). In exams, factor the expression first.
Step 2
Why this answer is correct
The correct answer is A. (-336). (\alpha-2 \beta+\alpha\beta-2 =\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=12\) and \(\alpha\beta=-28\), so the value is (-336). In exams, factor the expression first.
Step 3
Exam Tip
(\alpha-2 \beta+\alpha\beta-2 =\alpha\beta\(\alpha+\beta\)), जहां \(\alpha+\beta=12\) और \(\alpha\beta=-28\), इसलिए मान (-336) है। परीक्षा में अभिव्यक्ति को पहले factor करें।
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Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(x^2+14x+10=0\) के मूल क्या हैं?
What are the roots of \(x^2+14x+10=0\)?
#quadratic
#completing-square
#roots
A \(x=-7\pm\sqrt{39}\)
B \(x=7\pm\sqrt{39}\)
C \(x=-14\pm\sqrt{39}\)
D \(x=-7\pm39\)
Explanation opens after your attempt
Correct Answer
A. \(x=-7\pm\sqrt{39}\)
Step 1
Concept
Since ((x+7)2 =39), \(x=-7\pm\sqrt{39}\). In exams, write both values using \(\pm\).
Step 2
Why this answer is correct
The correct answer is A. \(x=-7\pm\sqrt{39}\). Since ((x+7)2 =39), \(x=-7\pm\sqrt{39}\). In exams, write both values using \(\pm\).
Step 3
Exam Tip
((x+7)2 =39), इसलिए \(x=-7\pm\sqrt{39}\) है। परीक्षा में \(\pm\) के दोनों मान लिखें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2+14x+10=0\), तो पूर्ण वर्ग विधि से सही रूप कौनसा है?
If \(x^2+14x+10=0\), which form is correct by completing square?
#quadratic
#completing-square
#irrational-roots
A ((x+7)2 =39)
B ((x-7)2 =39)
C ((x+14)2 =10)
D ((x+7)2 =10)
Explanation opens after your attempt
Correct Answer
A. ((x+7)2 =39)
Step 1
Concept
Adding (49) to \(x^2+14x=-10\) gives ((x+7)2 =39). In exams, add the same number to both sides.
Step 2
Why this answer is correct
The correct answer is A. ((x+7)2 =39). Adding (49) to \(x^2+14x=-10\) gives ((x+7)2 =39). In exams, add the same number to both sides.
Step 3
Exam Tip
\(x^2+14x=-10\) में (49) जोड़ने पर ((x+7)2 =39) मिलता है। परीक्षा में दोनों पक्षों में समान संख्या जोड़ें।
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Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(8x^2-14x-15=0\) के मूल क्या हैं?
What are the roots of \(8x^2-14x-15=0\)?
#quadratic
#roots
#factorisation
A \(x=\frac{5}{2},-\frac{3}{4}\)
B \(x=-\frac{5}{2},\frac{3}{4}\)
C (x=2,-5)
D \(x=\frac{14}{8},-1\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{5}{2},-\frac{3}{4}\)
Step 1
Concept
((4x+3)(2x-5)=0), so \(x=-\frac{3}{4}\) and \(\frac{5}{2}\). In exams, change signs while writing roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{2},-\frac{3}{4}\). ((4x+3)(2x-5)=0), so \(x=-\frac{3}{4}\) and \(\frac{5}{2}\). In exams, change signs while writing roots.
Step 3
Exam Tip
((4x+3)(2x-5)=0), इसलिए \(x=-\frac{3}{4}\) और \(\frac{5}{2}\) हैं। परीक्षा में संकेत बदलकर मूल लिखें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(8x^2-14x-15=0\) को हल करने में कौनसा गुणनखंड रूप सही है?
Which factorised form is correct for solving \(8x^2-14x-15=0\)?
#quadratic
#factorisation
#verification
A ((4x+3)(2x-5)=0)
B ((4x-3)(2x+5)=0)
C ((8x+5)(x-3)=0)
D ((x+5)(8x-3)=0)
Explanation opens after your attempt
Correct Answer
A. ((4x+3)(2x-5)=0)
Step 1
Concept
((4x+3)(2x-5)=8x-2 -20x+6x-15=8x-2 -14x-15), so it is correct. In exams, verify factorisation by expanding.
Step 2
Why this answer is correct
The correct answer is A. ((4x+3)(2x-5)=0). ((4x+3)(2x-5)=8x-2 -20x+6x-15=8x-2 -14x-15), so it is correct. In exams, verify factorisation by expanding.
Step 3
Exam Tip
((4x+3)(2x-5)=8x-2 -20x+6x-15=8x-2 -14x-15), इसलिए यह सही है। परीक्षा में गुणनखंड को विस्तार करके जांचें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(8x^2-23x-15=0\) का सही गुणनखंड रूप कौनसा है?
What is the correct factorised form of \(8x^2-23x-15=0\)?
#quadratic
#factorisation
#error-checking
A ((8x+5)(x-3)=0)
B ((8x-5)(x+3)=0)
C ((x+5)(8x-3)=0)
D ((8x-3)(x+5)=0)
Explanation opens after your attempt
Correct Answer
C. ((x+5)(8x-3)=0)
Step 1
Concept
((x+5)(8x-3)) does not expand to the given equation, so the options must be checked carefully. The correct factorisation is not present among careless options.
Step 2
Why this answer is correct
The correct answer is C. ((x+5)(8x-3)=0). ((x+5)(8x-3)) does not expand to the given equation, so the options must be checked carefully. The correct factorisation is not present among careless options.
Step 3
Exam Tip
((x+5)(8x-3)=8x-2 +37x-15) नहीं बनता; इसलिए विकल्पों में भी सावधानी चाहिए। सही गुणनखंड ((8x+5)(x-3)) नहीं है, अतः यह प्रश्न जाँच आधारित है।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(8x^2-23x-15=0\) को हल करने में कौनसा गुणनखंड रूप सही है?
Which factorised form is correct for solving \(8x^2-23x-15=0\)?
#quadratic
#factorisation
#audit
A ((8x+5)(x-3)=0)
B ((8x-5)(x+3)=0)
C ((x+5)(8x-3)=0)
D ((8x-3)(x+5)=0)
Explanation opens after your attempt
Correct Answer
A. ((8x+5)(x-3)=0)
Step 1
Concept
((8x+5)(x-3)=8x-2 -19x-15), so it is not for the given equation. In exams, verify each option by expansion.
Step 2
Why this answer is correct
The correct answer is A. ((8x+5)(x-3)=0). ((8x+5)(x-3)=8x-2 -19x-15), so it is not for the given equation. In exams, verify each option by expansion.
Step 3
Exam Tip
((8x+5)(x-3)=8x-2 -19x-15) नहीं बल्कि यह विस्तार गलत होगा; सही गुणनखंड ((8x+5)(x-3)) से (-24x+5x=-19x) बनता है। परीक्षा में विस्तार से हर विकल्प जांचें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2-16x+n=0\) के दो अलग वास्तविक मूल हैं, तो (n) के लिए कौनसी शर्त सही है?
If \(x^2-16x+n=0\) has two distinct real roots, which condition on (n) is correct?
#quadratic
#discriminant
#distinct-roots
A (n<64)
B (n>64)
C (n=64)
D \(n\ge64\)
Explanation opens after your attempt
Step 1
Concept
For two distinct real roots, (D>0), so (256-4n>0) and (n<64). In exams, connect (D>0) with distinct real roots.
Step 2
Why this answer is correct
The correct answer is A. (n<64). For two distinct real roots, (D>0), so (256-4n>0) and (n<64). In exams, connect (D>0) with distinct real roots.
Step 3
Exam Tip
दो अलग वास्तविक मूलों के लिए (D>0), इसलिए (256-4n>0) और (n<64) है। परीक्षा में (D>0) को अलग वास्तविक मूल से जोड़ें।
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Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2-16x+n=0\) के वास्तविक मूल नहीं हैं, तो (n) के लिए कौनसी शर्त सही है?
If \(x^2-16x+n=0\) has no real roots, which condition on (n) is correct?
#quadratic
#discriminant
#inequality
A (n>64)
B (n<64)
C (n=64)
D \(n\le64\)
Explanation opens after your attempt
Step 1
Concept
For no real roots, (D<0), so (256-4n<0) and (n>64). In exams, connect (D<0) with no real roots.
Step 2
Why this answer is correct
The correct answer is A. (n>64). For no real roots, (D<0), so (256-4n<0) and (n>64). In exams, connect (D<0) with no real roots.
Step 3
Exam Tip
वास्तविक मूल नहीं होने के लिए (D<0), इसलिए (256-4n<0) और (n>64) है। परीक्षा में (D<0) को वास्तविक मूल नहीं से जोड़ें।
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Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(8x^2-31x+15=0\) के मूल \(\alpha,\beta\) हैं, तो (\(\alpha-\beta\)2 ) क्या होगा?
If \(\alpha,\beta\) are roots of \(8x^2-31x+15=0\), what is (\(\alpha-\beta\)2 )?
#quadratic
#roots-expression
#identity
A \( \frac{481}{64}\)
B \( \frac{961}{64}\)
C \( \frac{15}{2}\)
D \( \frac{225}{64}\)
Explanation opens after your attempt
Correct Answer
A. \( \frac{481}{64}\)
Step 1
Concept
(\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta=\left\(\frac{31}{8}\right\)2 -\frac{15}{2}=\frac{481}{64}). In exams, convert fractions to a common denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{481}{64}\). (\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta=\left\(\frac{31}{8}\right\)2 -\frac{15}{2}=\frac{481}{64}). In exams, convert fractions to a common denominator.
Step 3
Exam Tip
(\(\alpha-\beta\)2 =\(\alpha+\beta\)2 -4\alpha\beta=\left\(\frac{31}{8}\right\)2 -\frac{15}{2}=\frac{481}{64}) है। परीक्षा में भिन्नों को समान हर में बदलें।
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Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(\alpha,\beta\) समीकरण \(8x^2-31x+15=0\) के मूल हैं, तो \(\alpha\beta\) क्या है?
If \(\alpha,\beta\) are roots of \(8x^2-31x+15=0\), what is \(\alpha\beta\)?
#quadratic
#product-of-roots
#formula
A \( \frac{15}{8}\)
B \( \frac{31}{8}\)
C (15)
D \(-\frac{15}{8}\)
Explanation opens after your attempt
Correct Answer
A. \( \frac{15}{8}\)
Step 1
Concept
The product of roots is \(\frac{c}{a}=\frac{15}{8}\). In exams, use \(\frac{c}{a}\) for the product.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{15}{8}\). The product of roots is \(\frac{c}{a}=\frac{15}{8}\). In exams, use \(\frac{c}{a}\) for the product.
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}=\frac{15}{8}\) है। परीक्षा में गुणनफल के लिए \(\frac{c}{a}\) का प्रयोग करें।
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Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(\alpha,\beta\) समीकरण \(8x^2-31x+15=0\) के मूल हैं, तो \(\alpha+\beta\) क्या है?
If \(\alpha,\beta\) are roots of \(8x^2-31x+15=0\), what is \(\alpha+\beta\)?
#quadratic
#sum-of-roots
#formula
A \( \frac{31}{8}\)
B \( \frac{15}{8}\)
C \( -\frac{31}{8}\)
D (8)
Explanation opens after your attempt
Correct Answer
A. \( \frac{31}{8}\)
Step 1
Concept
The sum of roots is \(-\frac{b}{a}=-\frac{-31}{8}=\frac{31}{8}\). In exams, keep the sign of (b) carefully.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{31}{8}\). The sum of roots is \(-\frac{b}{a}=-\frac{-31}{8}=\frac{31}{8}\). In exams, keep the sign of (b) carefully.
Step 3
Exam Tip
मूलों का योग \(-\frac{b}{a}=-\frac{-31}{8}=\frac{31}{8}\) है। परीक्षा में (b) का चिन्ह ध्यान से रखें।
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Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2-24x+135=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) क्या होगा?
If the roots of \(x^2-24x+135=0\) are \(\alpha\) and \(\beta\), what is \(\frac{1}{\alpha}+\frac{1}{\beta}\)?
#quadratic
#roots-expression
#reciprocal
A \( \frac{8}{45}\)
B \( \frac{45}{8}\)
C (24)
D (135)
Explanation opens after your attempt
Correct Answer
A. \( \frac{8}{45}\)
Step 1
Concept
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{24}{135}=\frac{8}{45}\). In exams, write the answer in simplest form.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{8}{45}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{24}{135}=\frac{8}{45}\). In exams, write the answer in simplest form.
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{24}{135}=\frac{8}{45}\) होता है। परीक्षा में उत्तर को सरल रूप में लिखें।
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Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2-27x+180=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो \(\alpha^2+\beta^2\) क्या होगा?
If the roots of \(x^2-27x+180=0\) are \(\alpha\) and \(\beta\), what is \(\alpha^2+\beta^2\)?
#quadratic
#roots-expression
#expert
A (369)
B (729)
C (180)
D (549)
Explanation opens after your attempt
Step 1
Concept
\(\alpha+\beta=27\) and \(\alpha\beta=180\), so (\alpha-2 +\beta-2 =272 -2(180)=369). In exams, remember (\(\alpha+\beta\)2 -2\alpha\beta).
Step 2
Why this answer is correct
The correct answer is A. (369). \(\alpha+\beta=27\) and \(\alpha\beta=180\), so (\alpha-2 +\beta-2 =272 -2(180)=369). In exams, remember (\(\alpha+\beta\)2 -2\alpha\beta).
Step 3
Exam Tip
\(\alpha+\beta=27\) और \(\alpha\beta=180\), इसलिए (\alpha-2 +\beta-2 =272 -2(180)=369) है। परीक्षा में (\(\alpha+\beta\)2 -2\alpha\beta) याद रखें।
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Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(10x^2-23x+p=0\) के मूलों का गुणनफल \(\frac{2}{5}\) है, तो (p) क्या होगा?
If the product of roots of \(10x^2-23x+p=0\) is \(\frac{2}{5}\), what is (p)?
#quadratic
#product-of-roots
#parameter
A (4)
B (10)
C \( \frac{1}{25}\)
D \( \frac{23}{5}\)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{p}{10}\), so \(\frac{p}{10}=\frac{2}{5}\) gives (p=4). In exams, use the product formula \(\frac{c}{a}\).
Step 2
Why this answer is correct
The correct answer is A. (4). The product of roots is \(\frac{p}{10}\), so \(\frac{p}{10}=\frac{2}{5}\) gives (p=4). In exams, use the product formula \(\frac{c}{a}\).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{p}{10}\) है, इसलिए \(\frac{p}{10}=\frac{2}{5}\) से (p=4) है। परीक्षा में गुणनफल का सूत्र \(\frac{c}{a}\) लगाएं।
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Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(9x^2+px+36=0\) के मूलों का योग (-10) है, तो (p) क्या होगा?
If the sum of roots of \(9x^2+px+36=0\) is (-10), what is (p)?
#quadratic
#sum-of-roots
#parameter
A (90)
B (-90)
C (10)
D (324)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is \(-\frac{p}{9}\), so \(-\frac{p}{9}=-10\) gives (p=90). In exams, remember the sum formula \(-\frac{b}{a}\).
Step 2
Why this answer is correct
The correct answer is A. (90). The sum of roots is \(-\frac{p}{9}\), so \(-\frac{p}{9}=-10\) gives (p=90). In exams, remember the sum formula \(-\frac{b}{a}\).
Step 3
Exam Tip
मूलों का योग \(-\frac{p}{9}\) है, इसलिए \(-\frac{p}{9}=-10\) से (p=90) है। परीक्षा में योग का सूत्र \(-\frac{b}{a}\) याद रखें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(x^2-14x+13=0\) के मूल द्विघात सूत्र से क्या हैं?
What are the roots of \(x^2-14x+13=0\) by quadratic formula?
#quadratic
#quadratic-formula
#perfect-square-discriminant
A (x=1,13)
B (x=-1,-13)
C \(x=7\pm6\)
D \(x=14\pm\sqrt{13}\)
Explanation opens after your attempt
Correct Answer
A. (x=1,13)
Step 1
Concept
(D=(-14)2 -4(1)(13)=144), so \(x=\frac{14\pm12}{2}\) gives (1) and (13). In exams, if (D) is a perfect square, simplify quickly.
Step 2
Why this answer is correct
The correct answer is A. (x=1,13). (D=(-14)2 -4(1)(13)=144), so \(x=\frac{14\pm12}{2}\) gives (1) and (13). In exams, if (D) is a perfect square, simplify quickly.
Step 3
Exam Tip
(D=(-14)2 -4(1)(13)=144), इसलिए \(x=\frac{14\pm12}{2}\) से (1) और (13) मिलते हैं। परीक्षा में (D) पूर्ण वर्ग हो तो उत्तर जल्दी सरल करें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(\frac{x+6}{x}=\frac{49}{x+6}\), \(x\neq0,-6\), के हल क्या हैं?
What are the solutions of \(\frac{x+6}{x}=\frac{49}{x+6}\), \(x\neq0,-6\)?
#quadratic
#rational-equation
#solutions
A (x=1,36)
B (x=-1,-36)
C (x=6,49)
D (x=37,36)
Explanation opens after your attempt
Correct Answer
A. (x=1,36)
Step 1
Concept
(x-2 -37x+36=(x-1)(x-36)), so (x=1) and (x=36). In exams, check solutions against excluded denominator values.
Step 2
Why this answer is correct
The correct answer is A. (x=1,36). (x-2 -37x+36=(x-1)(x-36)), so (x=1) and (x=36). In exams, check solutions against excluded denominator values.
Step 3
Exam Tip
(x-2 -37x+36=(x-1)(x-36)), इसलिए (x=1) और (x=36) हैं। परीक्षा में हर के निषिद्ध मानों से हलों की जांच करें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(\frac{x+6}{x}=\frac{49}{x+6}\), \(x\neq0,-6\), का सही द्विघात रूप कौनसा है?
What is the correct quadratic form of \(\frac{x+6}{x}=\frac{49}{x+6}\), \(x\neq0,-6\)?
#quadratic
#rational-equation
#standard-form
A \(x^2-37x+36=0\)
B \(x^2+12x-13=0\)
C \(x^2-49x+36=0\)
D \(x^2+37x+36=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-37x+36=0\)
Step 1
Concept
Cross multiplication gives ((x+6)2 =49x), so \(x^2+12x+36-49x=0\), and \(x^2-37x+36=0\). In exams, cross multiply carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-37x+36=0\). Cross multiplication gives ((x+6)2 =49x), so \(x^2+12x+36-49x=0\), and \(x^2-37x+36=0\). In exams, cross multiply carefully.
Step 3
Exam Tip
क्रॉस गुणा करने पर ((x+6)2 =49x), इसलिए \(x^2+12x+36-49x=0\) और \(x^2-37x+36=0\) है। परीक्षा में क्रॉस गुणा सावधानी से करें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(\frac{1}{x}+x=\frac{50}{7}\), \(x\neq0\), के हल क्या हैं?
What are the solutions of \(\frac{1}{x}+x=\frac{50}{7}\), \(x\neq0\)?
#quadratic
#reciprocal-equation
#solutions
A \(x=7,\frac{1}{7}\)
B \(x=-7,-\frac{1}{7}\)
C (x=50,7)
D \(x=\frac{50}{7},1\)
Explanation opens after your attempt
Correct Answer
A. \(x=7,\frac{1}{7}\)
Step 1
Concept
(7x-2 -50x+7=(7x-1)(x-7)), so \(x=\frac{1}{7}\) and (7). In exams, check whether obtained roots are valid in the original equation.
Step 2
Why this answer is correct
The correct answer is A. \(x=7,\frac{1}{7}\). (7x-2 -50x+7=(7x-1)(x-7)), so \(x=\frac{1}{7}\) and (7). In exams, check whether obtained roots are valid in the original equation.
Step 3
Exam Tip
(7x-2 -50x+7=(7x-1)(x-7)), इसलिए \(x=\frac{1}{7}\) और (7) हैं। परीक्षा में प्राप्त हल मूल समीकरण में मान्य हैं या नहीं जांचें।
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