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transformed_roots MCQ Questions for Class 10

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22 questions tagged with transformed_roots.

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 \(\alpha+6,\beta+6\)?

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 \(\alpha+5,\beta+5\)?

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 \(\alpha+4,\beta+4\)?

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 \(\alpha+3,\beta+3\)?

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 \(\alpha+2,\beta+2\)?

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 \(\alpha+1,\beta+1\)?

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 \(2\beta+3\)?

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 \(\beta-3\)?

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 \(3\beta-2\)?

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)?

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 \(\alpha+1,\beta+1\)?

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 \(\beta+3\)?

Explanation opens after your attempt
Correct Answer

A. (12)

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?

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?

Explanation opens after your attempt
Correct Answer

A. (-24)

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?

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 \(\beta+2\)?

Explanation opens after your attempt
Correct Answer

A. (9)

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?

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?

Explanation opens after your attempt
Correct Answer

A. (-15)

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?

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?

Explanation opens after your attempt
Correct Answer

A. (12)

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?

Explanation opens after your attempt
Correct Answer

A. (12)

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?

Explanation opens after your attempt
Correct Answer

A. (10)

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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