Question 1/7
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि \(x^2+ax+b=0\) की जड़ें \(\frac{1}{4+\sqrt{3}}\) और \(\frac{1}{4-\sqrt{3}}\) हैं, तो (a) का मान क्या है?
If the roots of \(x^2+ax+b=0\) are \(\frac{1}{4+\sqrt{3}}\) and \(\frac{1}{4-\sqrt{3}}\), what is the value of (a)?
#quadratic-roots
#rationalisation
#surd-roots
A -\(\frac{8}{13}\)
B \(\frac{8}{13}\)
C -\(\frac{4}{13}\)
D \(\frac{4}{13}\)
Explanation opens after your attempt
Correct Answer
A. -\(\frac{8}{13}\)
Step 1
Concept
The sum of roots is \(\frac{1}{4+\sqrt{3}}+\frac{1}{4-\sqrt{3}}=\frac{8}{13}\). In \(x^2+ax+b=0\), the sum is (-a), so \(a=-\frac{8}{13}\).
Step 2
Why this answer is correct
The correct answer is A. -\(\frac{8}{13}\). The sum of roots is \(\frac{1}{4+\sqrt{3}}+\frac{1}{4-\sqrt{3}}=\frac{8}{13}\). In \(x^2+ax+b=0\), the sum is (-a), so \(a=-\frac{8}{13}\).
Step 3
Exam Tip
जड़ों का योग \(\frac{1}{4+\sqrt{3}}+\frac{1}{4-\sqrt{3}}=\frac{8}{13}\) है। \(x^2+ax+b=0\) में योग (-a) होता है, इसलिए \(a=-\frac{8}{13}\)।
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Question 2/7
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि \(x^2+ax+b=0\) की जड़ें \(4+\sqrt{7}\) और \(4-\sqrt{7}\) हैं, तो (a+b) का मान क्या है?
If the roots of \(x^2+ax+b=0\) are \(4+\sqrt{7}\) and \(4-\sqrt{7}\), what is (a+b)?
#quadratic-roots
#surd-roots
#coefficients
A (0)
B (1)
C (2)
D (3)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (8), so (a=-8). The product is (9), so (b=9), hence (a+b=1).
Step 2
Why this answer is correct
The correct answer is B. (1). The sum of roots is (8), so (a=-8). The product is (9), so (b=9), hence (a+b=1).
Step 3
Exam Tip
जड़ों का योग (8) है, इसलिए (a=-8)। गुणनफल (9) है, इसलिए (b=9), अतः (a+b=1)।
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Question 3/7
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(x^2+ax+b=0\) की जड़ें \(\frac{1}{3+\sqrt{5}}\) और \(\frac{1}{3-\sqrt{5}}\) हैं, तो (a) का मान क्या है?
If the roots of \(x^2+ax+b=0\) are \(\frac{1}{3+\sqrt{5}}\) and \(\frac{1}{3-\sqrt{5}}\), what is the value of (a)?
#quadratic-roots
#rationalisation
#surd-roots
A \(-\frac{3}{2}\)
B \(\frac{3}{2}\)
C (-3)
D (3)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{3}{2}\)
Step 1
Concept
After rationalising, the sum of roots is \(\frac{3}{2}\). In \(x^2+ax+b=0\), the sum is (-a), so \(a=-\frac{3}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{3}{2}\). After rationalising, the sum of roots is \(\frac{3}{2}\). In \(x^2+ax+b=0\), the sum is (-a), so \(a=-\frac{3}{2}\).
Step 3
Exam Tip
रैशनलाइज करने पर जड़ों का योग \(\frac{3}{2}\) मिलता है। \(x^2+ax+b=0\) में जड़ों का योग (-a) होता है, इसलिए \(a=-\frac{3}{2}\)।
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Question 4/7
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(x^2+ax+b=0\) की जड़ें \(3+\sqrt{2}\) और \(3-\sqrt{2}\) हैं, तो (a+b) का मान क्या है?
If the roots of \(x^2+ax+b=0\) are \(3+\sqrt{2}\) and \(3-\sqrt{2}\), what is (a+b)?
#quadratic-roots
#surd-roots
#coefficients
A (-1)
B (1)
C (7)
D (13)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (6), so (a=-6). The product is (7), so (b=7), hence (a+b=1).
Step 2
Why this answer is correct
The correct answer is B. (1). The sum of roots is (6), so (a=-6). The product is (7), so (b=7), hence (a+b=1).
Step 3
Exam Tip
जड़ों का योग (6) है, इसलिए (a=-6)। गुणनफल (7) है, इसलिए (b=7), अतः (a+b=1)।
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Question 5/7
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि \(x^2+ax+b=0\) की जड़ें \(\frac{1}{2+\sqrt{3}}\) और \(\frac{1}{2-\sqrt{3}}\) हैं, तो (a) का मान क्या है?
If the roots of \(x^2+ax+b=0\) are \(\frac{1}{2+\sqrt{3}}\) and \(\frac{1}{2-\sqrt{3}}\), what is the value of (a)?
#quadratic-roots
#rationalisation
#surd-roots
A (-4)
B (4)
C (-2)
D (2)
Explanation opens after your attempt
Step 1
Concept
The given roots become \(2-\sqrt{3}\) and \(2+\sqrt{3}\). Their sum is (4), so (a=-4).
Step 2
Why this answer is correct
The correct answer is A. (-4). The given roots become \(2-\sqrt{3}\) and \(2+\sqrt{3}\). Their sum is (4), so (a=-4).
Step 3
Exam Tip
दी गई जड़ें \(2-\sqrt{3}\) और \(2+\sqrt{3}\) बनती हैं। उनका योग (4) है, इसलिए (a=-4)।
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Question 6/7
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि \(x^2+ax+b=0\) की जड़ें \(2+\sqrt{3}\) और \(2-\sqrt{3}\) हैं, तो (a+b) का मान क्या है?
If the roots of \(x^2+ax+b=0\) are \(2+\sqrt{3}\) and \(2-\sqrt{3}\), what is (a+b)?
#quadratic-roots
#surd-roots
#coefficients
A (-3)
B (-1)
C (1)
D (3)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (4), so (a=-4). The product is (1), so (b=1), hence (a+b=-3).
Step 2
Why this answer is correct
The correct answer is A. (-3). The sum of roots is (4), so (a=-4). The product is (1), so (b=1), hence (a+b=-3).
Step 3
Exam Tip
जड़ों का योग (4) है, इसलिए (a=-4)। गुणनफल (1) है, इसलिए (b=1), अतः (a+b=-3)।
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Question 7/7
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि किसी द्विघात समीकरण की जड़ें \(2+\sqrt{5}\) और \(2-\sqrt{5}\) हैं, तो समीकरण कौन-सा है?
If the roots of a quadratic equation are \(2+\sqrt{5}\) and \(2-\sqrt{5}\), which is the equation?
#quadratic-roots
#forming-equation
#surd-roots
A \(x^2-4x-1=0\)
B \(x^2+4x-1=0\)
C \(x^2-4x+1=0\)
D \(x^2+4x+1=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-4x-1=0\)
Step 1
Concept
\(The sum of roots is (4) and the product is (-1). Use (x^2-(\)sum)x+product\(=0) to get the answer.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-4x-1=0). The sum of roots is (4) and the product is (-1). Use (x^2-(\)sum)x+product\(=0) to get the answer.\)
Step 3
Exam Tip
जड़ों का योग (4) और गुणनफल (-1) है। \(समीकरण (x^2-(\)योग)x+गुणनफल\(=0) से उत्तर मिलता है\)।
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