Class 10 repeated root MCQ Questions

Question 1/34 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

\(x^2-2\sqrt{17}x+17=0\) का मूल क्या है?

What is the root of \(x^2-2\sqrt{17}x+17=0\)?

Explanation opens after your attempt

Correct Answer

A. \(x=\sqrt{17}\)

Step 1

Concept

(\(x-\sqrt{17}\)²=0), so the repeated root is \(\sqrt{17}\). In exams, ((x-a)²=0) gives (x=a).

Step 2

Why this answer is correct

The correct answer is A. \(x=\sqrt{17}\). (\(x-\sqrt{17}\)²=0), so the repeated root is \(\sqrt{17}\). In exams, ((x-a)²=0) gives (x=a).

Step 3

Exam Tip

(\(x-\sqrt{17}\)²=0), इसलिए दोहराया हुआ मूल \(\sqrt{17}\) है। परीक्षा में ((x-a)²=0) से (x=a) मिलता है।

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Question 2/34 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

\(x^2+2\sqrt{13}x+13=0\) का मूल क्या है?

What is the root of \(x^2+2\sqrt{13}x+13=0\)?

Explanation opens after your attempt

Correct Answer

A. \(x=-\sqrt{13}\)

Step 1

Concept

(\(x+\sqrt{13}\)²=0), so the repeated root is \(-\sqrt{13}\). In exams, ((x+a)²=0) gives (x=-a).

Step 2

Why this answer is correct

The correct answer is A. \(x=-\sqrt{13}\). (\(x+\sqrt{13}\)²=0), so the repeated root is \(-\sqrt{13}\). In exams, ((x+a)²=0) gives (x=-a).

Step 3

Exam Tip

(\(x+\sqrt{13}\)²=0), इसलिए दोहराया हुआ मूल \(-\sqrt{13}\) है। परीक्षा में ((x+a)²=0) से (x=-a) मिलता है।

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Question 3/34 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

\(x^2-2\sqrt{11}x+11=0\) का मूल क्या है?

What is the root of \(x^2-2\sqrt{11}x+11=0\)?

Explanation opens after your attempt

Correct Answer

A. \(x=\sqrt{11}\)

Step 1

Concept

(\(x-\sqrt{11}\)²=0), so the repeated root is \(\sqrt{11}\). In exams, ((x-a)²=0) gives (x=a).

Step 2

Why this answer is correct

The correct answer is A. \(x=\sqrt{11}\). (\(x-\sqrt{11}\)²=0), so the repeated root is \(\sqrt{11}\). In exams, ((x-a)²=0) gives (x=a).

Step 3

Exam Tip

(\(x-\sqrt{11}\)²=0), इसलिए दोहराया हुआ मूल \(\sqrt{11}\) है। परीक्षा में ((x-a)²=0) से (x=a) मिलता है।

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Question 4/34 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

\(x^2+2\sqrt{7}x+7=0\) का मूल क्या है?

What is the root of \(x^2+2\sqrt{7}x+7=0\)?

Explanation opens after your attempt

Correct Answer

A. \(x=-\sqrt{7}\)

Step 1

Concept

(\(x+\sqrt{7}\)²=0), so the repeated root is \(-\sqrt{7}\). In exams, ((x+a)²=0) gives (x=-a).

Step 2

Why this answer is correct

The correct answer is A. \(x=-\sqrt{7}\). (\(x+\sqrt{7}\)²=0), so the repeated root is \(-\sqrt{7}\). In exams, ((x+a)²=0) gives (x=-a).

Step 3

Exam Tip

(\(x+\sqrt{7}\)²=0), इसलिए दोहराया हुआ मूल \(-\sqrt{7}\) है। परीक्षा में ((x+a)²=0) से (x=-a) मिलता है।

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Question 5/34 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

\(x^2-2\sqrt{5}x+5=0\) का मूल क्या है?

What is the root of \(x^2-2\sqrt{5}x+5=0\)?

Explanation opens after your attempt

Correct Answer

A. \(x=\sqrt{5}\)

Step 1

Concept

(\(x-\sqrt{5}\)²=0), so the repeated root is \(\sqrt{5}\). In exams, ((x-a)²=0) gives (x=a).

Step 2

Why this answer is correct

The correct answer is A. \(x=\sqrt{5}\). (\(x-\sqrt{5}\)²=0), so the repeated root is \(\sqrt{5}\). In exams, ((x-a)²=0) gives (x=a).

Step 3

Exam Tip

(\(x-\sqrt{5}\)²=0), इसलिए दोहराया हुआ मूल \(\sqrt{5}\) है। परीक्षा में ((x-a)²=0) से (x=a) मिलता है।

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Question 6/34 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

\(x^2+2\sqrt{3}x+3=0\) का मूल क्या है?

What is the root of \(x^2+2\sqrt{3}x+3=0\)?

Explanation opens after your attempt

Correct Answer

A. \(x=-\sqrt{3}\)

Step 1

Concept

(\(x+\sqrt{3}\)²=0), so the repeated root is \(-\sqrt{3}\). In exams, ((x+a)²=0) gives (x=-a).

Step 2

Why this answer is correct

The correct answer is A. \(x=-\sqrt{3}\). (\(x+\sqrt{3}\)²=0), so the repeated root is \(-\sqrt{3}\). In exams, ((x+a)²=0) gives (x=-a).

Step 3

Exam Tip

(\(x+\sqrt{3}\)²=0), इसलिए दोहराया हुआ मूल \(-\sqrt{3}\) है। परीक्षा में ((x+a)²=0) से (x=-a) मिलता है।

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Question 7/34 Medium Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

\(49x^2-42x+9=0\) का मूल क्या है?

What is the root of \(49x^2-42x+9=0\)?

Explanation opens after your attempt

Correct Answer

A. \(x=\frac{3}{7}\)

Step 1

Concept

((7x-3)²=0), so (7x-3=0) and \(x=\frac{3}{7}\). In exams, solve the linear equation after square form.

Step 2

Why this answer is correct

The correct answer is A. \(x=\frac{3}{7}\). ((7x-3)²=0), so (7x-3=0) and \(x=\frac{3}{7}\). In exams, solve the linear equation after square form.

Step 3

Exam Tip

((7x-3)²=0), इसलिए (7x-3=0) और \(x=\frac{3}{7}\) है। परीक्षा में वर्ग रूप के बाद रैखिक समीकरण हल करें।

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Question 8/34 Medium Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

\(36x^2-60x+25=0\) का दोहराया हुआ मूल क्या है?

What is the repeated root of \(36x^2-60x+25=0\)?

Explanation opens after your attempt

Correct Answer

A. \(x=\frac{5}{6}\)

Step 1

Concept

((6x-5)²=0), so (6x-5=0) and \(x=\frac{5}{6}\). In exams, write the repeated root as a correct fraction.

Step 2

Why this answer is correct

The correct answer is A. \(x=\frac{5}{6}\). ((6x-5)²=0), so (6x-5=0) and \(x=\frac{5}{6}\). In exams, write the repeated root as a correct fraction.

Step 3

Exam Tip

((6x-5)²=0), इसलिए (6x-5=0) और \(x=\frac{5}{6}\) है। परीक्षा में दोहराए हुए मूल को सही भिन्न में लिखें।

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Question 9/34 Medium Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

\(25x^2-20x+4=0\) का मूल क्या है?

What is the root of \(25x^2-20x+4=0\)?

Explanation opens after your attempt

Correct Answer

A. \(x=\frac{2}{5}\)

Step 1

Concept

((5x-2)²=0), so (5x-2=0) and \(x=\frac{2}{5}\). In exams, solve the linear equation after square form.

Step 2

Why this answer is correct

The correct answer is A. \(x=\frac{2}{5}\). ((5x-2)²=0), so (5x-2=0) and \(x=\frac{2}{5}\). In exams, solve the linear equation after square form.

Step 3

Exam Tip

((5x-2)²=0), इसलिए (5x-2=0) और \(x=\frac{2}{5}\) है। परीक्षा में वर्ग रूप के बाद रैखिक समीकरण हल करें।

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Question 10/34 Medium Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

\(16x^2-24x+9=0\) का दोहराया हुआ मूल क्या है?

What is the repeated root of \(16x^2-24x+9=0\)?

Explanation opens after your attempt

Correct Answer

A. \(x=\frac{3}{4}\)

Step 1

Concept

((4x-3)²=0), so (4x-3=0) and \(x=\frac{3}{4}\). In exams, write the repeated root as a correct fraction.

Step 2

Why this answer is correct

The correct answer is A. \(x=\frac{3}{4}\). ((4x-3)²=0), so (4x-3=0) and \(x=\frac{3}{4}\). In exams, write the repeated root as a correct fraction.

Step 3

Exam Tip

((4x-3)²=0), इसलिए (4x-3=0) और \(x=\frac{3}{4}\) है। परीक्षा में दोहराए हुए मूल को भी सही भिन्न में लिखें।

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Question 11/34 Medium Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

\(9x^2-30x+25=0\) का मूल क्या है?

What is the root of \(9x^2-30x+25=0\)?

Explanation opens after your attempt

Correct Answer

A. \(x=\frac{5}{3}\)

Step 1

Concept

((3x-5)²=0), so (3x-5=0) and \(x=\frac{5}{3}\). In exams, solve the linear equation after square form.

Step 2

Why this answer is correct

The correct answer is A. \(x=\frac{5}{3}\). ((3x-5)²=0), so (3x-5=0) and \(x=\frac{5}{3}\). In exams, solve the linear equation after square form.

Step 3

Exam Tip

((3x-5)²=0), इसलिए (3x-5=0) और \(x=\frac{5}{3}\) है। परीक्षा में वर्ग रूप के बाद रैखिक समीकरण हल करें।

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Question 12/34 Medium Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

\(4x^2-12x+9=0\) का दोहराया हुआ मूल क्या है?

What is the repeated root of \(4x^2-12x+9=0\)?

Explanation opens after your attempt

Correct Answer

A. \(x=\frac{3}{2}\)

Step 1

Concept

((2x-3)²=0), so (2x-3=0) and \(x=\frac{3}{2}\). In exams, write the repeated root with the correct value.

Step 2

Why this answer is correct

The correct answer is A. \(x=\frac{3}{2}\). ((2x-3)²=0), so (2x-3=0) and \(x=\frac{3}{2}\). In exams, write the repeated root with the correct value.

Step 3

Exam Tip

((2x-3)²=0), इसलिए (2x-3=0) और \(x=\frac{3}{2}\) है। परीक्षा में दोहराए हुए मूल को भी सही मान से लिखें।

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Question 13/34 Easy Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

\(x^2+18x+81=0\) का दोहराया हुआ मूल क्या है?

What is the repeated root of \(x^2+18x+81=0\)?

Explanation opens after your attempt

Correct Answer

A. (x=-9)

Step 1

Concept

((x+9)²=0), so the repeated root is (-9). In exams, a perfect square equation has equal roots.

Step 2

Why this answer is correct

The correct answer is A. (x=-9). ((x+9)²=0), so the repeated root is (-9). In exams, a perfect square equation has equal roots.

Step 3

Exam Tip

((x+9)²=0), इसलिए दोहराया हुआ मूल (-9) है। परीक्षा में पूर्ण वर्ग समीकरण में दोनों मूल समान होते हैं।

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Question 14/34 Easy Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

\(x^2+14x+49=0\) का दोहराया हुआ मूल क्या है?

What is the repeated root of \(x^2+14x+49=0\)?

Explanation opens after your attempt

Correct Answer

A. (x=-7)

Step 1

Concept

((x+7)²=0), so the repeated root is (-7). In exams, ((x+a)²=0) gives (x=-a).

Step 2

Why this answer is correct

The correct answer is A. (x=-7). ((x+7)²=0), so the repeated root is (-7). In exams, ((x+a)²=0) gives (x=-a).

Step 3

Exam Tip

((x+7)²=0), इसलिए दोहराया हुआ मूल (-7) है। परीक्षा में ((x+a)²=0) से (x=-a) मिलता है।

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Question 15/34 Easy Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

\(x^2-12x+36=0\) का मूल क्या होगा?

What will be the root of \(x^2-12x+36=0\)?

Explanation opens after your attempt

Correct Answer

A. (x=6)

Step 1

Concept

((x-6)²=0), so both equal roots are (x=6). In exams, a perfect square equation gives a repeated root.

Step 2

Why this answer is correct

The correct answer is A. (x=6). ((x-6)²=0), so both equal roots are (x=6). In exams, a perfect square equation gives a repeated root.

Step 3

Exam Tip

((x-6)²=0), इसलिए दोनों समान मूल (x=6) हैं। परीक्षा में पूर्ण वर्ग समीकरण में दोहराया हुआ मूल मिलता है।

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Question 16/34 Easy Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

\(x^2-6x+9=0\) का दोहराया हुआ मूल क्या है?

What is the repeated root of \(x^2-6x+9=0\)?

Explanation opens after your attempt

Correct Answer

A. (x=3)

Step 1

Concept

(x^-2-6x+9=(x-3)²), so the repeated root is (3). In exams, a perfect square gives equal roots.

Step 2

Why this answer is correct

The correct answer is A. (x=3). (x^-2-6x+9=(x-3)²), so the repeated root is (3). In exams, a perfect square gives equal roots.

Step 3

Exam Tip

(x^-2-6x+9=(x-3)²), इसलिए दोहराया हुआ मूल (3) है। परीक्षा में पूर्ण वर्ग में दोनों मूल समान होते हैं।

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Question 17/34 Medium Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33

समीकरण \(x^2-22x+121=0\) के मूलों का योग और गुणनफल क्रमशः क्या हैं?

For \(x^2-22x+121=0\), what are the sum and product of roots respectively?

Explanation opens after your attempt

Correct Answer

A. (22) और (121)(22) and (121)

Step 1

Concept

It is ((x-11)²=0), so both roots are (11). The sum is (22) and the product is (121).

Step 2

Why this answer is correct

The correct answer is A. (22) और (121) / (22) and (121). It is ((x-11)²=0), so both roots are (11). The sum is (22) and the product is (121).

Step 3

Exam Tip

यह ((x-11)²=0) है इसलिए दोनों मूल (11) हैं। योग (22) और गुणनफल (121) होगा।

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Question 18/34 Medium Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33

समीकरण \(25x^2-10x+1=0\) का दोहराया हुआ मूल क्या है?

What is the repeated root of \(25x^2-10x+1=0\)?

Explanation opens after your attempt

Correct Answer

A. \(\frac{1}{5}\)

Step 1

Concept

(25x^-2-10x+1=(5x-1)²). Therefore the repeated root is \(\frac{1}{5}\).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{1}{5}\). (25x^-2-10x+1=(5x-1)²). Therefore the repeated root is \(\frac{1}{5}\).

Step 3

Exam Tip

(25x^-2-10x+1=(5x-1)²) है। इसलिए दोहराया मूल \(\frac{1}{5}\) है।

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Question 19/34 Medium Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33

यदि किसी द्विघात समीकरण में (a=1), (b=-20), (c=100) है तो मूल कौन से होंगे?

If a quadratic equation has (a=1), (b=-20), and (c=100), what will be the roots?

Explanation opens after your attempt

Correct Answer

A. (10) और (10)(10) and (10)

Step 1

Concept

The equation is \(x^2-20x+100=0\), which becomes ((x-10)²=0). Therefore both roots are (10).

Step 2

Why this answer is correct

The correct answer is A. (10) और (10) / (10) and (10). The equation is \(x^2-20x+100=0\), which becomes ((x-10)²=0). Therefore both roots are (10).

Step 3

Exam Tip

समीकरण \(x^2-20x+100=0\) है जो ((x-10)²=0) बनता है। इसलिए दोनों मूल (10) हैं।

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Question 20/34 Medium Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32

समीकरण \(x^2-18x+81=0\) के मूलों का योग और गुणनफल क्रमशः क्या हैं?

For \(x^2-18x+81=0\), what are the sum and product of roots respectively?

Explanation opens after your attempt

Correct Answer

A. (18) और (81)(18) and (81)

Step 1

Concept

It is ((x-9)²=0), so both roots are (9). The sum is (18) and the product is (81).

Step 2

Why this answer is correct

The correct answer is A. (18) और (81) / (18) and (81). It is ((x-9)²=0), so both roots are (9). The sum is (18) and the product is (81).

Step 3

Exam Tip

यह ((x-9)²=0) है इसलिए दोनों मूल (9) हैं। योग (18) और गुणनफल (81) होगा।

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Question 21/34 Medium Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32

समीकरण \(16x^2-8x+1=0\) का दोहराया हुआ मूल क्या है?

What is the repeated root of \(16x^2-8x+1=0\)?

Explanation opens after your attempt

Correct Answer

A. \(\frac{1}{4}\)

Step 1

Concept

(16x^-2-8x+1=(4x-1)²). Therefore the repeated root is \(\frac{1}{4}\).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{1}{4}\). (16x^-2-8x+1=(4x-1)²). Therefore the repeated root is \(\frac{1}{4}\).

Step 3

Exam Tip

(16x^-2-8x+1=(4x-1)²) है। इसलिए दोहराया मूल \(\frac{1}{4}\) है।

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Question 22/34 Medium Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32

यदि किसी द्विघात समीकरण में (a=1), (b=-16), (c=64) है तो मूल कौन से होंगे?

If a quadratic equation has (a=1), (b=-16), and (c=64), what will be the roots?

Explanation opens after your attempt

Correct Answer

A. (8) और (8)(8) and (8)

Step 1

Concept

The equation is \(x^2-16x+64=0\), which becomes ((x-8)²=0). Therefore both roots are (8).

Step 2

Why this answer is correct

The correct answer is A. (8) और (8) / (8) and (8). The equation is \(x^2-16x+64=0\), which becomes ((x-8)²=0). Therefore both roots are (8).

Step 3

Exam Tip

समीकरण \(x^2-16x+64=0\) है जो ((x-8)²=0) बनता है। इसलिए दोनों मूल (8) हैं।

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Question 23/34 Medium Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31

समीकरण \(x^2-14x+49=0\) के मूलों का योग और गुणनफल क्रमशः क्या हैं?

For \(x^2-14x+49=0\), what are the sum and product of roots respectively?

Explanation opens after your attempt

Correct Answer

A. (14) और (49)(14) and (49)

Step 1

Concept

It is ((x-7)²=0), so both roots are (7). The sum is (14) and the product is (49).

Step 2

Why this answer is correct

The correct answer is A. (14) और (49) / (14) and (49). It is ((x-7)²=0), so both roots are (7). The sum is (14) and the product is (49).

Step 3

Exam Tip

यह ((x-7)²=0) है इसलिए दोनों मूल (7) हैं। योग (14) और गुणनफल (49) होगा।

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Question 24/34 Medium Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31

समीकरण \(9x^2-6x+1=0\) का दोहराया हुआ मूल क्या है?

What is the repeated root of \(9x^2-6x+1=0\)?

Explanation opens after your attempt

Correct Answer

A. \(\frac{1}{3}\)

Step 1

Concept

(9x^-2-6x+1=(3x-1)²). Therefore the repeated root is \(\frac{1}{3}\).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{1}{3}\). (9x^-2-6x+1=(3x-1)²). Therefore the repeated root is \(\frac{1}{3}\).

Step 3

Exam Tip

(9x^-2-6x+1=(3x-1)²) है। इसलिए दोहराया मूल \(\frac{1}{3}\) है।

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Question 25/34 Medium Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31

यदि किसी द्विघात समीकरण में (a=1), (b=-10), (c=25) है तो मूल कौन से होंगे?

If a quadratic equation has (a=1), (b=-10), and (c=25), what will be the roots?

Explanation opens after your attempt

Correct Answer

A. (5) और (5)(5) and (5)

Step 1

Concept

The equation is \(x^2-10x+25=0\), which becomes ((x-5)²=0). Therefore both roots are (5).

Step 2

Why this answer is correct

The correct answer is A. (5) और (5) / (5) and (5). The equation is \(x^2-10x+25=0\), which becomes ((x-5)²=0). Therefore both roots are (5).

Step 3

Exam Tip

समीकरण \(x^2-10x+25=0\) है जो ((x-5)²=0) बनता है। इसलिए दोनों मूल (5) हैं।

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Question 26/34 Easy Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33

समीकरण \(x^2-18x+81=0\) का मूल कौन सा है?

Which is the root of \(x^2-18x+81=0\)?

Explanation opens after your attempt

Correct Answer

A. (9) दो बार(9) twice

Step 1

Concept

(x^-2-18x+81=(x-9)²). Therefore the repeated root is (9).

Step 2

Why this answer is correct

The correct answer is A. (9) दो बार / (9) twice. (x^-2-18x+81=(x-9)²). Therefore the repeated root is (9).

Step 3

Exam Tip

(x^-2-18x+81=(x-9)²) है। इसलिए दोहराया मूल (9) है।

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Question 27/34 Easy Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33

समीकरण \(x^2+16x+64=0\) के मूल कौन से हैं?

What are the roots of \(x^2+16x+64=0\)?

Explanation opens after your attempt

Correct Answer

B. (-8) और (-8)(-8) and (-8)

Step 1

Concept

(x^-2+16x+64=(x+8)²). Therefore both roots are (-8).

Step 2

Why this answer is correct

The correct answer is B. (-8) और (-8) / (-8) and (-8). (x^-2+16x+64=(x+8)²). Therefore both roots are (-8).

Step 3

Exam Tip

(x^-2+16x+64=(x+8)²) है। इसलिए दोनों मूल (-8) हैं।

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Question 28/34 Easy Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33

समीकरण \(x^2-12x+36=0\) का दोहराया हुआ मूल क्या है?

What is the repeated root of \(x^2-12x+36=0\)?

Explanation opens after your attempt

Correct Answer

A. (6)

Step 1

Concept

(x^-2-12x+36=(x-6)²). Therefore the repeated root is (6).

Step 2

Why this answer is correct

The correct answer is A. (6). (x^-2-12x+36=(x-6)²). Therefore the repeated root is (6).

Step 3

Exam Tip

(x^-2-12x+36=(x-6)²) है। इसलिए दोहराया हुआ मूल (6) है।

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Question 29/34 Easy Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32

समीकरण \(x^2+14x+49=0\) का मूल कौन सा है?

Which is the root of \(x^2+14x+49=0\)?

Explanation opens after your attempt

Correct Answer

B. (-7) दो बार(-7) twice

Step 1

Concept

(x^-2+14x+49=(x+7)²). Therefore the repeated root is (-7).

Step 2

Why this answer is correct

The correct answer is B. (-7) दो बार / (-7) twice. (x^-2+14x+49=(x+7)²). Therefore the repeated root is (-7).

Step 3

Exam Tip

(x^-2+14x+49=(x+7)²) है। इसलिए दोहराया मूल (-7) है।

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Question 30/34 Easy Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32

समीकरण \(x^2+12x+36=0\) के मूल कौन से हैं?

What are the roots of \(x^2+12x+36=0\)?

Explanation opens after your attempt

Correct Answer

B. (-6) और (-6)(-6) and (-6)

Step 1

Concept

(x^-2+12x+36=(x+6)²). Therefore both roots are (-6).

Step 2

Why this answer is correct

The correct answer is B. (-6) और (-6) / (-6) and (-6). (x^-2+12x+36=(x+6)²). Therefore both roots are (-6).

Step 3

Exam Tip

(x^-2+12x+36=(x+6)²) है। इसलिए दोनों मूल (-6) हैं।

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repeated_root MCQ Questions for Class 10

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यदि किसी द्विघात समीकरण में (a=1), (b=-20), (c=100) है तो मूल कौन से होंगे?

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समीकरण \(x^2-18x+81=0\) के मूलों का योग और गुणनफल क्रमशः क्या हैं?

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