Question 1/22
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 2/22
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35
यदि \(x^2-14x+33=0\) के मूल \(\alpha,\beta\) हैं, तो नए मूल \(\alpha+5,\beta+5\) वाला समीकरण कौनसा है?
If roots of \(x^2-14x+33=0\) are \(\alpha,\beta\), which equation has roots \(\alpha+5,\beta+5\)?
#quadratic
#transformed-roots
#equation
A \(x^2-24x+128=0\)
B \(x^2-14x+33=0\)
C \(x^2-4x+8=0\)
D \(x^2+24x+128=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-24x+128=0\)
Step 1
Concept
The roots are (3,11), so new roots are (8,16), and the equation is ((x-8)(x-16)=0). In exams, form the new roots and then the new equation.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-24x+128=0\). The roots are (3,11), so new roots are (8,16), and the equation is ((x-8)(x-16)=0). In exams, form the new roots and then the new equation.
Step 3
Exam Tip
मूल (3,11) हैं, इसलिए नए मूल (8,16) होंगे और समीकरण ((x-8)(x-16)=0) है। परीक्षा में नए मूल बनाकर नया समीकरण लिखें।
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Question 3/22
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34
यदि \(x^2-12x+20=0\) के मूल \(\alpha,\beta\) हैं, तो नए मूल \(\alpha+4,\beta+4\) वाला समीकरण कौनसा है?
If roots of \(x^2-12x+20=0\) are \(\alpha,\beta\), which equation has roots \(\alpha+4,\beta+4\)?
#quadratic
#transformed-roots
#equation
A \(x^2-20x+84=0\)
B \(x^2-12x+20=0\)
C \(x^2-4x+4=0\)
D \(x^2+20x+84=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-20x+84=0\)
Step 1
Concept
The roots are (2,10), so new roots are (6,14), and the equation is ((x-6)(x-14)=0). In exams, form the new roots and then the new equation.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-20x+84=0\). The roots are (2,10), so new roots are (6,14), and the equation is ((x-6)(x-14)=0). In exams, form the new roots and then the new equation.
Step 3
Exam Tip
मूल (2,10) हैं, इसलिए नए मूल (6,14) होंगे और समीकरण ((x-6)(x-14)=0) है। परीक्षा में नए मूल बनाकर नया समीकरण लिखें।
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Question 4/22
Hard Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2-10x+16=0\) के मूल \(\alpha,\beta\) हैं, तो नए मूल \(\alpha+3,\beta+3\) वाला समीकरण कौनसा है?
If roots of \(x^2-10x+16=0\) are \(\alpha,\beta\), which equation has roots \(\alpha+3,\beta+3\)?
#quadratic
#transformed-roots
#equation
A \(x^2-16x+55=0\)
B \(x^2-10x+16=0\)
C \(x^2-4x+7=0\)
D \(x^2+16x+55=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-16x+55=0\)
Step 1
Concept
The roots are (2,8), so new roots are (5,11), and the equation is ((x-5)(x-11)=0). In exams, form the new roots and then the new equation.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-16x+55=0\). The roots are (2,8), so new roots are (5,11), and the equation is ((x-5)(x-11)=0). In exams, form the new roots and then the new equation.
Step 3
Exam Tip
मूल (2,8) हैं, इसलिए नए मूल (5,11) होंगे और समीकरण ((x-5)(x-11)=0) है। परीक्षा में नए मूल बनाकर नया समीकरण लिखें।
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Question 5/22
Hard Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35
यदि \(x^2-8x+12=0\) के मूल \(\alpha,\beta\) हैं, तो नए मूल \(\alpha+2,\beta+2\) वाला समीकरण कौनसा है?
If roots of \(x^2-8x+12=0\) are \(\alpha,\beta\), which equation has roots \(\alpha+2,\beta+2\)?
#quadratic
#transformed-roots
#equation
A \(x^2-12x+32=0\)
B \(x^2-8x+12=0\)
C \(x^2-4x+8=0\)
D \(x^2+12x+32=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-12x+32=0\)
Step 1
Concept
The roots are (2,6), so new roots are (4,8), and the equation is ((x-4)(x-8)=0). In exams, form the new roots and then the new equation.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-12x+32=0\). The roots are (2,6), so new roots are (4,8), and the equation is ((x-4)(x-8)=0). In exams, form the new roots and then the new equation.
Step 3
Exam Tip
मूल (2,6) हैं, इसलिए नए मूल (4,8) होंगे और समीकरण ((x-4)(x-8)=0) है। परीक्षा में नए मूल बनाकर नया समीकरण लिखें।
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Question 6/22
Hard Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34
यदि \(x^2-6x+5=0\) के मूल \(\alpha,\beta\) हैं, तो नए मूल \(\alpha+1,\beta+1\) वाला समीकरण कौनसा है?
If roots of \(x^2-6x+5=0\) are \(\alpha,\beta\), which equation has roots \(\alpha+1,\beta+1\)?
#quadratic
#transformed-roots
#equation
A \(x^2-8x+12=0\)
B \(x^2-6x+5=0\)
C \(x^2-4x+4=0\)
D \(x^2+8x+12=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-8x+12=0\)
Step 1
Concept
The roots are (1,5), so new roots are (2,6), and the equation is ((x-2)(x-6)=0). In exams, form the new roots and then the new equation.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-8x+12=0\). The roots are (1,5), so new roots are (2,6), and the equation is ((x-2)(x-6)=0). In exams, form the new roots and then the new equation.
Step 3
Exam Tip
मूल (1,5) हैं, इसलिए नए मूल (2,6) होंगे और समीकरण ((x-2)(x-6)=0) है। परीक्षा में नए मूल बनाकर नया समीकरण लिखें।
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Question 7/22
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि \(x^2-7x+12=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(2\alpha+3\) और \(2\beta+3\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-7x+12=0\), which equation has roots \(2\alpha+3\) and \(2\beta+3\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-20x+99=0\)
B \(x^2-14x+99=0\)
C \(x^2-20x+91=0\)
D \(x^2+20x+99=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-20x+99=0\)
Step 1
Concept
The original roots are (3) and (4). The new roots are (9) and (11), so the equation is \(x^2-20x+99=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-20x+99=0\). The original roots are (3) and (4). The new roots are (9) and (11), so the equation is \(x^2-20x+99=0\).
Step 3
Exam Tip
मूल जड़ें (3) और (4) हैं। नई जड़ें (9) और (11) हैं, इसलिए समीकरण \(x^2-20x+99=0\) है।
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Question 8/22
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(x^2-9x+14=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha-3\) और \(\beta-3\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-9x+14=0\), which equation has roots \(\alpha-3\) and \(\beta-3\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-3x-4=0\)
B \(x^2+3x-4=0\)
C \(x^2-3x+4=0\)
D \(x^2-9x+14=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-3x-4=0\)
Step 1
Concept
The original roots are (2) and (7). The new roots are (-1) and (4), so the equation is \(x^2-3x-4=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-3x-4=0\). The original roots are (2) and (7). The new roots are (-1) and (4), so the equation is \(x^2-3x-4=0\).
Step 3
Exam Tip
मूल जड़ें (2) और (7) हैं। नई जड़ें (-1) और (4) होंगी, इसलिए समीकरण \(x^2-3x-4=0\) है।
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Question 9/22
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि \(x^2-6x+5=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(3\alpha-2\) और \(3\beta-2\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-6x+5=0\), which equation has roots \(3\alpha-2\) and \(3\beta-2\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-14x+13=0\)
B \(x^2-18x+45=0\)
C \(x^2-14x+25=0\)
D \(x^2+14x+13=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-14x+13=0\)
Step 1
Concept
The original roots are (1) and (5), so the new roots are (1) and (13). Their equation is \(x^2-14x+13=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-14x+13=0\). The original roots are (1) and (5), so the new roots are (1) and (13). Their equation is \(x^2-14x+13=0\).
Step 3
Exam Tip
मूल जड़ें (1) और (5) हैं, इसलिए नई जड़ें (1) और (13) हैं। उनका समीकरण \(x^2-14x+13=0\) है।
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Question 10/22
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि \(x^2+px+q=0\) की जड़ें \(\alpha,\beta\) हैं और \(\alpha+1,\beta+1\), \(x^2-5x+6=0\) की जड़ें हैं, तो (p,q) क्या होंगे?
If \(\alpha,\beta\) are the roots of \(x^2+px+q=0\) and \(\alpha+1,\beta+1\) are the roots of \(x^2-5x+6=0\), what are (p,q)?
#quadratic-roots
#transformed-roots
#parameter
A (p=-3,\ q=2)
B (p=3,\ q=2)
C (p=-2,\ q=3)
D (p=2,\ q=-3)
Explanation opens after your attempt
Correct Answer
A. (p=-3,\ q=2)
Step 1
Concept
The sum of new roots is \(\alpha+\beta+2=5\), so (p=-3). From product (q-p+1=6), we get (q=2).
Step 2
Why this answer is correct
The correct answer is A. (p=-3,\ q=2). The sum of new roots is \(\alpha+\beta+2=5\), so (p=-3). From product (q-p+1=6), we get (q=2).
Step 3
Exam Tip
नई जड़ों का योग \(\alpha+\beta+2=5\) है, इसलिए (p=-3)। गुणनफल (q-p+1=6) से (q=2) मिलता है।
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Question 11/22
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
\(x^2-5x+6=0\) की जड़ें \(\alpha,\beta\) हैं। \(\alpha+1,\beta+1\) जड़ों वाला समीकरण कौन-सा है?
The roots of \(x^2-5x+6=0\) are \(\alpha,\beta\). Which equation has roots \(\alpha+1,\beta+1\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-7x+12=0\)
B \(x^2-5x+12=0\)
C \(x^2-6x+8=0\)
D \(x^2+7x+12=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-7x+12=0\)
Step 1
Concept
The original roots are (2) and (3), so the new roots are (3) and (4). Their equation is \(x^2-7x+12=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-7x+12=0\). The original roots are (2) and (3), so the new roots are (3) and (4). Their equation is \(x^2-7x+12=0\).
Step 3
Exam Tip
मूल जड़ें (2) और (3) हैं, इसलिए नई जड़ें (3) और (4) होंगी। उनका समीकरण \(x^2-7x+12=0\) है।
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Question 12/22
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(x^2-6x+5=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(\alpha+3\) और \(\beta+3\) का योग क्या होगा?
If \(\alpha\) and \(\beta\) are roots of \(x^2-6x+5=0\), what is the sum of \(\alpha+3\) and \(\beta+3\)?
#roots
#transformed_roots
#sum
A (12)
B (6)
C (9)
D (15)
Explanation opens after your attempt
Step 1
Concept
The old sum of roots is (6). The new sum is (\(\alpha+3\)+\(\beta+3\)=6+6=12).
Step 2
Why this answer is correct
The correct answer is A. (12). The old sum of roots is (6). The new sum is (\(\alpha+3\)+\(\beta+3\)=6+6=12).
Step 3
Exam Tip
पुराने मूलों का योग (6) है। नया योग (\(\alpha+3\)+\(\beta+3\)=6+6=12) होगा।
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Question 13/22
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(x^2-4x+3=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(3\alpha\) और \(3\beta\) को मूल मानकर समीकरण कौन सा होगा?
If \(\alpha\) and \(\beta\) are roots of \(x^2-4x+3=0\), which equation has \(3\alpha\) and \(3\beta\) as roots?
#roots
#transformed_roots
#equation
A \(x^2-12x+27=0\)
B \(x^2-4x+27=0\)
C \(x^2-12x+9=0\)
D \(x^2+12x+27=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-12x+27=0\)
Step 1
Concept
The old sum is (4) and product is (3). The new sum is (12) and product is (27), so the equation is \(x^2-12x+27=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-12x+27=0\). The old sum is (4) and product is (3). The new sum is (12) and product is (27), so the equation is \(x^2-12x+27=0\).
Step 3
Exam Tip
पुराने योग (4) और गुणनफल (3) हैं। नए योग (12) और गुणनफल (27) होंगे इसलिए \(x^2-12x+27=0\) है।
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Question 14/22
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
समीकरण \(x^2-6x-16=0\) के मूलों को (2) घटाने पर नए मूलों का गुणनफल क्या होगा?
If each root of \(x^2-6x-16=0\) is decreased by (2), what is the product of the new roots?
#roots
#transformed_roots
#product
A (-24)
B (-16)
C (24)
D (16)
Explanation opens after your attempt
Step 1
Concept
The old roots are (8) and (-2). The new roots are (6) and (-4), so the product is (-24).
Step 2
Why this answer is correct
The correct answer is A. (-24). The old roots are (8) and (-2). The new roots are (6) and (-4), so the product is (-24).
Step 3
Exam Tip
पुराने मूल (8) और (-2) हैं। नए मूल (6) और (-4) होंगे इसलिए गुणनफल (-24) है।
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Question 15/22
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(x^2-6x-16=0\) के मूलों को (1) बढ़ा दिया जाए तो नए मूलों से बना मोनिक समीकरण कौन सा होगा?
If each root of \(x^2-6x-16=0\) is increased by (1), which monic equation is formed from the new roots?
#roots
#transformed_roots
#new_equation
A \(x^2-8x-9=0\)
B \(x^2-6x-16=0\)
C \(x^2-4x-9=0\)
D \(x^2+8x-9=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-8x-9=0\)
Step 1
Concept
The old roots are (8) and (-2). The new roots are (9) and (-1), so the equation is \(x^2-8x-9=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-8x-9=0\). The old roots are (8) and (-2). The new roots are (9) and (-1), so the equation is \(x^2-8x-9=0\).
Step 3
Exam Tip
पुराने मूल (8) और (-2) हैं। नए मूल (9) और (-1) होंगे इसलिए समीकरण \(x^2-8x-9=0\) है।
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Question 16/22
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि \(x^2-5x+4=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(\alpha+2\) और \(\beta+2\) का योग क्या होगा?
If \(\alpha\) and \(\beta\) are roots of \(x^2-5x+4=0\), what is the sum of \(\alpha+2\) and \(\beta+2\)?
#roots
#transformed_roots
#sum
A (9)
B (5)
C (7)
D (4)
Explanation opens after your attempt
Step 1
Concept
The old sum of roots is (5). The new sum is (\(\alpha+2\)+\(\beta+2\)=\alpha+\beta+4=9).
Step 2
Why this answer is correct
The correct answer is A. (9). The old sum of roots is (5). The new sum is (\(\alpha+2\)+\(\beta+2\)=\alpha+\beta+4=9).
Step 3
Exam Tip
पुराने मूलों का योग (5) है। नए योग (\(\alpha+2\)+\(\beta+2\)=\alpha+\beta+4=9) होगा।
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Question 17/22
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि \(x^2-3x+2=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(2\alpha\) और \(2\beta\) को मूल मानकर समीकरण कौन सा होगा?
If \(\alpha\) and \(\beta\) are roots of \(x^2-3x+2=0\), which equation has \(2\alpha\) and \(2\beta\) as roots?
#roots
#transformed_roots
#equation
A \(x^2-6x+8=0\)
B \(x^2-3x+8=0\)
C \(x^2-6x+4=0\)
D \(x^2+6x+8=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x+8=0\)
Step 1
Concept
The old sum is (3) and product is (2). The new sum is (6) and product is (8), so the equation is \(x^2-6x+8=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x+8=0\). The old sum is (3) and product is (2). The new sum is (6) and product is (8), so the equation is \(x^2-6x+8=0\).
Step 3
Exam Tip
पुराने योग (3) और गुणनफल (2) हैं। नए योग (6) और गुणनफल (8) होंगे इसलिए \(x^2-6x+8=0\) है।
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Question 18/22
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
समीकरण \(x^2-4x-12=0\) के मूलों को (1) घटाने पर नए मूलों का गुणनफल क्या होगा?
If each root of \(x^2-4x-12=0\) is decreased by (1), what is the product of the new roots?
#roots
#transformed_roots
#product
A (-15)
B (-12)
C (15)
D (12)
Explanation opens after your attempt
Step 1
Concept
The old roots are (6) and (-2). The new roots are (5) and (-3), so the product is (-15).
Step 2
Why this answer is correct
The correct answer is A. (-15). The old roots are (6) and (-2). The new roots are (5) and (-3), so the product is (-15).
Step 3
Exam Tip
पुराने मूल (6) और (-2) हैं। नए मूल (5) और (-3) होंगे इसलिए गुणनफल (-15) है।
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Question 19/22
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि \(x^2-4x-12=0\) के मूलों को (1) बढ़ा दिया जाए तो नए मूलों से बना मोनिक समीकरण कौन सा होगा?
If each root of \(x^2-4x-12=0\) is increased by (1), which monic equation is formed from the new roots?
#roots
#transformed_roots
#new_equation
A \(x^2-6x-15=0\)
B \(x^2-4x-12=0\)
C \(x^2-2x-15=0\)
D \(x^2+6x-15=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x-15=0\)
Step 1
Concept
The old roots are (6) and (-2). The new roots are (7) and (-1), so the equation is \(x^2-6x-7=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x-15=0\). The old roots are (6) and (-2). The new roots are (7) and (-1), so the equation is \(x^2-6x-7=0\).
Step 3
Exam Tip
पुराने मूल (6) और (-2) हैं। नए मूल (7) और (-1) होंगे इसलिए समीकरण \(x^2-6x-7=0\) नहीं बल्कि \(x^2-6x-7=0\) होता है।
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Question 20/22
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि \(4\alpha\) और \(4\beta\) नए मूल हैं तथा \(\alpha+\beta=3\) है तो नए मूलों का योग क्या होगा?
If \(4\alpha\) and \(4\beta\) are new roots and \(\alpha+\beta=3\), what is the sum of the new roots?
#roots
#transformed_roots
#sum
A (12)
B (7)
C (3)
D \(\frac{3}{4}\)
Explanation opens after your attempt
Step 1
Concept
The sum of new roots is (4\alpha+4\beta=4\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is also multiplied by that factor.
Step 2
Why this answer is correct
The correct answer is A. (12). The sum of new roots is (4\alpha+4\beta=4\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is also multiplied by that factor.
Step 3
Exam Tip
नए मूलों का योग (4\alpha+4\beta=4\(\alpha+\beta\)=12) है। गुणक लगे मूलों में योग भी उसी गुणक से गुणा होता है।
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Question 21/22
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(3\alpha\) और \(3\beta\) नए मूल हैं तथा \(\alpha+\beta=4\) है तो नए मूलों का योग क्या होगा?
If \(3\alpha\) and \(3\beta\) are new roots and \(\alpha+\beta=4\), what is the sum of the new roots?
#roots
#transformed_roots
#sum
A (12)
B (4)
C (7)
D \(\frac{4}{3}\)
Explanation opens after your attempt
Step 1
Concept
The sum of new roots is (3\alpha+3\beta=3\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is multiplied by the same factor.
Step 2
Why this answer is correct
The correct answer is A. (12). The sum of new roots is (3\alpha+3\beta=3\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is multiplied by the same factor.
Step 3
Exam Tip
नए मूलों का योग (3\alpha+3\beta=3\(\alpha+\beta\)=12) है। गुणक लगे मूलों में योग पर भी वही गुणक लगता है।
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Question 22/22
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि \(2\alpha\) और \(2\beta\) मूल हैं तथा \(\alpha+\beta=5\) है तो नए मूलों का योग क्या होगा?
If \(2\alpha\) and \(2\beta\) are roots and \(\alpha+\beta=5\), what is the sum of the new roots?
#roots
#transformed_roots
#sum
A (10)
B (5)
C (20)
D \(\frac{5}{2}\)
Explanation opens after your attempt
Step 1
Concept
The sum of new roots is (2\alpha+2\beta=2\(\alpha+\beta\)=10). When roots are multiplied by a factor, the sum is also multiplied by that factor.
Step 2
Why this answer is correct
The correct answer is A. (10). The sum of new roots is (2\alpha+2\beta=2\(\alpha+\beta\)=10). When roots are multiplied by a factor, the sum is also multiplied by that factor.
Step 3
Exam Tip
नए मूलों का योग (2\alpha+2\beta=2\(\alpha+\beta\)=10) है। गुणक लगे मूलों में योग पर भी वही गुणक लगता है।
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