निम्न में से किस समीकरण के मूल वास्तविक, अपरिमेय और असमान हैं?
Which of the following equations has real, irrational, and distinct roots?
#quadratic-equations
#choose-equation
#irrational-roots
A \(x^2-2\sqrt{2}x-1=0\)
B \(x^2-4x+4=0\)
C \(x^2+2x+5=0\)
D \(x^2-5x+6=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-2\sqrt{2}x-1=0\)
Step 1
Concept
In option (A), (D=\(-2\sqrt{2}\)2 -4(1)(-1)=12). (12) is positive but not a perfect square.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-2\sqrt{2}x-1=0\). In option (A), (D=\(-2\sqrt{2}\)2 -4(1)(-1)=12). (12) is positive but not a perfect square.
Step 3
Exam Tip
विकल्प (A) में (D=\(-2\sqrt{2}\)2 -4(1)(-1)=12) है। (12) धनात्मक है पर पूर्ण वर्ग नहीं है।
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निम्न में से किस समीकरण के मूल समान हैं?
Which of the following equations has equal roots?
#quadratic-equations
#choose-equation
#equal-roots
A \(7x^2-10\sqrt{7}x+25=0\)
B \(7x^2-9\sqrt{7}x+25=0\)
C \(7x^2-8\sqrt{7}x+25=0\)
D \(7x^2-6\sqrt{7}x+25=0\)
Explanation opens after your attempt
Correct Answer
A. \(7x^2-10\sqrt{7}x+25=0\)
Step 1
Concept
In option (A), (D=\(-10\sqrt{7}\)2 -4(7)(25)=0). Equal roots need (D=0).
Step 2
Why this answer is correct
The correct answer is A. \(7x^2-10\sqrt{7}x+25=0\). In option (A), (D=\(-10\sqrt{7}\)2 -4(7)(25)=0). Equal roots need (D=0).
Step 3
Exam Tip
विकल्प (A) में (D=\(-10\sqrt{7}\)2 -4(7)(25)=0) है। समान मूलों के लिए (D=0) चाहिए।
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निम्न में से किस समीकरण के दो वास्तविक परिमेय और असमान मूल हैं?
Which of the following equations has two real rational and distinct roots?
#quadratic-equations
#choose-equation
#rational-roots
A \(x^2-17x+72=0\)
B \(x^2-17x+80=0\)
C \(x^2+17x+80=0\) जहाँ (D=-31) / \(x^2+17x+80=0\) with (D=-31)
D \(x^2-18x+81=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-17x+72=0\)
Step 1
Concept
In option (A), (D=(-17)2 -4(1)(72)=1). A positive perfect-square (D) gives rational distinct roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-17x+72=0\). In option (A), (D=(-17)2 -4(1)(72)=1). A positive perfect-square (D) gives rational distinct roots.
Step 3
Exam Tip
विकल्प (A) में (D=(-17)2 -4(1)(72)=1) है। धनात्मक पूर्ण वर्ग (D) परिमेय असमान मूल देता है।
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निम्न में से किस समीकरण के दो वास्तविक और असमान मूल हैं?
Which of the following equations has two real and distinct roots?
#quadratic-equations
#choose-equation
#real-distinct-roots
A \(x^2-11x+18=0\)
B \(x^2+4x+4=0\)
C \(x^2+2x+6=0\)
D \(x^2-8x+16=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-11x+18=0\)
Step 1
Concept
In option (A), (D=(-11)2 -4(1)(18)=49). When (D>0), two distinct real roots exist.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-11x+18=0\). In option (A), (D=(-11)2 -4(1)(18)=49). When (D>0), two distinct real roots exist.
Step 3
Exam Tip
विकल्प (A) में (D=(-11)2 -4(1)(18)=49) है। (D>0) होने पर दो असमान वास्तविक मूल मिलते हैं।
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कौन सा समीकरण वास्तविक, अपरिमेय और भिन्न मूल देता है?
Which equation gives real, irrational and distinct roots?
#quadratic equations
#irrational distinct roots
#choose equation
A \(x^2-10x+23=0\)
B \(x^2-10x+24=0\)
C \(x^2-10x+25=0\)
D \(x^2+10x+26=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-10x+23=0\)
Step 1
Concept
In the first equation, (D=100-92=8>0), and (8) is not a perfect square. So the roots are real, irrational and distinct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x+23=0\). In the first equation, (D=100-92=8>0), and (8) is not a perfect square. So the roots are real, irrational and distinct.
Step 3
Exam Tip
पहले समीकरण में (D=100-92=8>0) है और (8) पूर्ण वर्ग नहीं है। इसलिए मूल वास्तविक, अपरिमेय और भिन्न हैं।
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निम्न में से किस समीकरण के कोई वास्तविक मूल नहीं हैं?
Which of the following equations has no real roots?
#quadratic-equations
#choose-equation
#no-real-roots
A \(x^2+2x+5=0\)
B \(x^2-2x-5=0\)
C \(x^2-5x+6=0\)
D \(x^2+4x+4=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+2x+5=0\)
Step 1
Concept
For option (A), (D=22 -4(1)(5)=-16). An equation with negative (D) has no real roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+2x+5=0\). For option (A), (D=22 -4(1)(5)=-16). An equation with negative (D) has no real roots.
Step 3
Exam Tip
विकल्प (A) के लिए (D=22 -4(1)(5)=-16) है। ऋणात्मक (D) वाले समीकरण में वास्तविक मूल नहीं होते।
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निम्न में से किस समीकरण के दो वास्तविक और समान मूल हैं?
Which of the following equations has two real and equal roots?
#quadratic-equations
#choose-equation
#equal-roots
A \(x^2-6x+9=0\)
B \(x^2-6x+8=0\)
C \(x^2+6x+10=0\)
D \(x^2-6x+5=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x+9=0\)
Step 1
Concept
In option (A), (D=(-6)2 -4(1)(9)=0). The option with (D=0) gives equal roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x+9=0\). In option (A), (D=(-6)2 -4(1)(9)=0). The option with (D=0) gives equal roots.
Step 3
Exam Tip
विकल्प (A) में (D=(-6)2 -4(1)(9)=0) है। (D=0) वाला विकल्प समान मूल देता है।
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कौन सा समीकरण वास्तविक, अपरिमेय और भिन्न मूल रखता है?
Which equation has real, irrational and distinct roots?
#quadratic equations
#choose equation
#irrational roots
A \(x^2-2x-3=0\)
B \(x^2-2x-2=0\)
C \(x^2-2x+1=0\)
D \(x^2+2x+5=0\)
Explanation opens after your attempt
Correct Answer
B. \(x^2-2x-2=0\)
Step 1
Concept
In the second equation (D=(-2)2 -4(1)(-2)=12). (12) is positive but not a perfect square.
Step 2
Why this answer is correct
The correct answer is B. \(x^2-2x-2=0\). In the second equation (D=(-2)2 -4(1)(-2)=12). (12) is positive but not a perfect square.
Step 3
Exam Tip
दूसरे समीकरण में (D=(-2)2 -4(1)(-2)=12) है। (12) धनात्मक है पर पूर्ण वर्ग नहीं है।
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कौन सा समीकरण वास्तविक, परिमेय और भिन्न मूल रखता है?
Which equation has real, rational and distinct roots?
#quadratic equations
#choose equation
#rational roots
A \(x^2-9x+20=0\)
B \(x^2-9x+21=0\)
C \(x^2+9x+30=0\)
D \(x^2+2x+5=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-9x+20=0\)
Step 1
Concept
In the first equation (D=(-9)2 -4(1)(20)=1). Hence the roots are real, rational and distinct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-9x+20=0\). In the first equation (D=(-9)2 -4(1)(20)=1). Hence the roots are real, rational and distinct.
Step 3
Exam Tip
पहले समीकरण में (D=(-9)2 -4(1)(20)=1) है। इसलिए मूल वास्तविक, परिमेय और भिन्न हैं।
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कौन सा समीकरण वास्तविक मूल नहीं रखता है?
Which equation has no real roots?
#quadratic equations
#choose equation
#no real roots
A \(x^2+2x+10=0\)
B \(x^2-2x+1=0\)
C \(x^2-2x-10=0\)
D \(x^2+10x+2=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+2x+10=0\)
Step 1
Concept
In the first equation (D=(2)2 -4(1)(10)=-36<0). Therefore there is no real root.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+2x+10=0\). In the first equation (D=(2)2 -4(1)(10)=-36<0). Therefore there is no real root.
Step 3
Exam Tip
पहले समीकरण में (D=(2)2 -4(1)(10)=-36<0) है। इसलिए कोई वास्तविक मूल नहीं है।
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कौन सा समीकरण वास्तविक और समान मूल रखता है?
Which equation has real and equal roots?
#quadratic equations
#choose equation
#equal roots
A \(x^2-8x+16=0\)
B \(x^2-8x+15=0\)
C \(x^2-8x+20=0\)
D \(x^2+8x+12=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-8x+16=0\)
Step 1
Concept
In the first equation (D=(-8)2 -4(1)(16)=0). So its roots are real and equal.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-8x+16=0\). In the first equation (D=(-8)2 -4(1)(16)=0). So its roots are real and equal.
Step 3
Exam Tip
पहले समीकरण में (D=(-8)2 -4(1)(16)=0) है। इसलिए इसके मूल वास्तविक और समान हैं।
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कौन सा समीकरण वास्तविक और भिन्न मूल रखता है?
Which equation has real and distinct roots?
#quadratic equations
#choose equation
#D positive
A \(x^2-3x+2=0\)
B \(x^2+2x+1=0\)
C \(x^2+x+1=0\)
D \(4x^2+4x+1=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-3x+2=0\)
Step 1
Concept
In the first equation (D=(-3)2 -4(1)(2)=1>0). Hence its roots are real and distinct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-3x+2=0\). In the first equation (D=(-3)2 -4(1)(2)=1>0). Hence its roots are real and distinct.
Step 3
Exam Tip
पहले समीकरण में (D=(-3)2 -4(1)(2)=1>0) है। इसलिए उसके मूल वास्तविक और भिन्न हैं।
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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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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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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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