Concept-wise Practice

roots-expression MCQ Questions for Class 10

roots-expression se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.

Practice Questions

30 questions tagged with roots-expression.

Question 1/30 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

यदि \(x^2-26x+165=0\) के मूल \(\alpha,\beta\) हैं, तो \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) क्या होगा?

If the roots of \(x^2-26x+165=0\) are \(\alpha,\beta\), what is \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{346}{165}\)

Step 1

Concept

\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=26\) and \(\alpha\beta=165\), so the value is \(\frac{676-330}{165}=\frac{346}{165}\). In exams, convert expressions into sum and product.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{346}{165}\). \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=26\) and \(\alpha\beta=165\), so the value is \(\frac{676-330}{165}=\frac{346}{165}\). In exams, convert expressions into sum and product.

Step 3

Exam Tip

\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), जहां \(\alpha+\beta=26\) और \(\alpha\beta=165\), इसलिए मान \(\frac{676-330}{165}=\frac{346}{165}\) है। परीक्षा में अभिव्यक्ति को योग और गुणनफल में बदलें।

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Question 2/30 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

यदि \(x^2-12x-28=0\) के मूल \(\alpha,\beta\) हैं, तो \(\alpha^2\beta+\alpha\beta^2\) क्या होगा?

If the roots of \(x^2-12x-28=0\) are \(\alpha,\beta\), what is \(\alpha^2\beta+\alpha\beta^2\)?

Explanation opens after your attempt
Correct Answer

A. (-336)

Step 1

Concept

(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=12\) and \(\alpha\beta=-28\), so the value is (-336). In exams, factor the expression first.

Step 2

Why this answer is correct

The correct answer is A. (-336). (\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=12\) and \(\alpha\beta=-28\), so the value is (-336). In exams, factor the expression first.

Step 3

Exam Tip

(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), जहां \(\alpha+\beta=12\) और \(\alpha\beta=-28\), इसलिए मान (-336) है। परीक्षा में अभिव्यक्ति को पहले factor करें।

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Question 3/30 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

यदि \(8x^2-31x+15=0\) के मूल \(\alpha,\beta\) हैं, तो (\(\alpha-\beta\)2) क्या होगा?

If \(\alpha,\beta\) are roots of \(8x^2-31x+15=0\), what is (\(\alpha-\beta\)2)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{481}{64}\)

Step 1

Concept

(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{31}{8}\right\)2-\frac{15}{2}=\frac{481}{64}). In exams, convert fractions to a common denominator.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{481}{64}\). (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{31}{8}\right\)2-\frac{15}{2}=\frac{481}{64}). In exams, convert fractions to a common denominator.

Step 3

Exam Tip

(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{31}{8}\right\)2-\frac{15}{2}=\frac{481}{64}) है। परीक्षा में भिन्नों को समान हर में बदलें।

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Question 4/30 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

यदि \(x^2-24x+135=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) क्या होगा?

If the roots of \(x^2-24x+135=0\) are \(\alpha\) and \(\beta\), what is \(\frac{1}{\alpha}+\frac{1}{\beta}\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{8}{45}\)

Step 1

Concept

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{24}{135}=\frac{8}{45}\). In exams, write the answer in simplest form.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{8}{45}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{24}{135}=\frac{8}{45}\). In exams, write the answer in simplest form.

Step 3

Exam Tip

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{24}{135}=\frac{8}{45}\) होता है। परीक्षा में उत्तर को सरल रूप में लिखें।

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Question 5/30 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

यदि \(x^2-27x+180=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो \(\alpha^2+\beta^2\) क्या होगा?

If the roots of \(x^2-27x+180=0\) are \(\alpha\) and \(\beta\), what is \(\alpha^2+\beta^2\)?

Explanation opens after your attempt
Correct Answer

A. (369)

Step 1

Concept

\(\alpha+\beta=27\) and \(\alpha\beta=180\), so (\alpha-2+\beta-2=272-2(180)=369). In exams, remember (\(\alpha+\beta\)2-2\alpha\beta).

Step 2

Why this answer is correct

The correct answer is A. (369). \(\alpha+\beta=27\) and \(\alpha\beta=180\), so (\alpha-2+\beta-2=272-2(180)=369). In exams, remember (\(\alpha+\beta\)2-2\alpha\beta).

Step 3

Exam Tip

\(\alpha+\beta=27\) और \(\alpha\beta=180\), इसलिए (\alpha-2+\beta-2=272-2(180)=369) है। परीक्षा में (\(\alpha+\beta\)2-2\alpha\beta) याद रखें।

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Question 6/30 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

यदि \(x^2-22x+117=0\) के मूल \(\alpha,\beta\) हैं, तो \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) क्या होगा?

If the roots of \(x^2-22x+117=0\) are \(\alpha,\beta\), what is \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{250}{117}\)

Step 1

Concept

\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=22\) and \(\alpha\beta=117\), so the value is \(\frac{484-234}{117}=\frac{250}{117}\). In exams, convert expressions into sum and product.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{250}{117}\). \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=22\) and \(\alpha\beta=117\), so the value is \(\frac{484-234}{117}=\frac{250}{117}\). In exams, convert expressions into sum and product.

Step 3

Exam Tip

\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), जहां \(\alpha+\beta=22\) और \(\alpha\beta=117\), इसलिए मान \(\frac{484-234}{117}=\frac{250}{117}\) है। परीक्षा में अभिव्यक्ति को योग और गुणनफल में बदलें।

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Question 7/30 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

यदि \(x^2-10x-24=0\) के मूल \(\alpha,\beta\) हैं, तो \(\alpha^2\beta+\alpha\beta^2\) क्या होगा?

If the roots of \(x^2-10x-24=0\) are \(\alpha,\beta\), what is \(\alpha^2\beta+\alpha\beta^2\)?

Explanation opens after your attempt
Correct Answer

A. (-240)

Step 1

Concept

(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=10\) and \(\alpha\beta=-24\), so the value is (-240). In exams, factor the expression first.

Step 2

Why this answer is correct

The correct answer is A. (-240). (\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=10\) and \(\alpha\beta=-24\), so the value is (-240). In exams, factor the expression first.

Step 3

Exam Tip

(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), जहां \(\alpha+\beta=10\) और \(\alpha\beta=-24\), इसलिए मान (-240) है। परीक्षा में अभिव्यक्ति को पहले factor करें।

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Question 8/30 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

यदि \(7x^2-25x+12=0\) के मूल \(\alpha,\beta\) हैं, तो (\(\alpha-\beta\)2) क्या होगा?

If \(\alpha,\beta\) are roots of \(7x^2-25x+12=0\), what is (\(\alpha-\beta\)2)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{289}{49}\)

Step 1

Concept

(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{25}{7}\right\)2-\frac{48}{7}=\frac{289}{49}). In exams, convert fractions to a common denominator.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{289}{49}\). (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{25}{7}\right\)2-\frac{48}{7}=\frac{289}{49}). In exams, convert fractions to a common denominator.

Step 3

Exam Tip

(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{25}{7}\right\)2-\frac{48}{7}=\frac{289}{49}) है। परीक्षा में भिन्नों को समान हर में बदलें।

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Question 9/30 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

यदि \(x^2-20x+91=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) क्या होगा?

If the roots of \(x^2-20x+91=0\) are \(\alpha\) and \(\beta\), what is \(\frac{1}{\alpha}+\frac{1}{\beta}\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{20}{91}\)

Step 1

Concept

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{20}{91}\). In exams, first write sum and product in reciprocal questions.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{20}{91}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{20}{91}\). In exams, first write sum and product in reciprocal questions.

Step 3

Exam Tip

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{20}{91}\) होता है। परीक्षा में reciprocal वाले प्रश्न में पहले योग और गुणनफल लिखें।

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Question 10/30 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

यदि \(x^2-23x+126=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो \(\alpha^2+\beta^2\) क्या होगा?

If the roots of \(x^2-23x+126=0\) are \(\alpha\) and \(\beta\), what is \(\alpha^2+\beta^2\)?

Explanation opens after your attempt
Correct Answer

A. (277)

Step 1

Concept

\(\alpha+\beta=23\) and \(\alpha\beta=126\), so (\alpha-2+\beta-2=232-2(126)=277). In exams, remember (\(\alpha+\beta\)2-2\alpha\beta).

Step 2

Why this answer is correct

The correct answer is A. (277). \(\alpha+\beta=23\) and \(\alpha\beta=126\), so (\alpha-2+\beta-2=232-2(126)=277). In exams, remember (\(\alpha+\beta\)2-2\alpha\beta).

Step 3

Exam Tip

\(\alpha+\beta=23\) और \(\alpha\beta=126\), इसलिए (\alpha-2+\beta-2=232-2(126)=277) है। परीक्षा में (\(\alpha+\beta\)2-2\alpha\beta) याद रखें।

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Question 11/30 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

यदि \(x^2-18x+77=0\) के मूल \(\alpha,\beta\) हैं, तो \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) क्या होगा?

If the roots of \(x^2-18x+77=0\) are \(\alpha,\beta\), what is \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{170}{77}\)

Step 1

Concept

\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=18\) and \(\alpha\beta=77\), so the value is \(\frac{324-154}{77}=\frac{170}{77}\). In exams, convert expressions into sum and product.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{170}{77}\). \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=18\) and \(\alpha\beta=77\), so the value is \(\frac{324-154}{77}=\frac{170}{77}\). In exams, convert expressions into sum and product.

Step 3

Exam Tip

\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), जहां \(\alpha+\beta=18\) और \(\alpha\beta=77\), इसलिए मान \(\frac{324-154}{77}=\frac{170}{77}\) है। परीक्षा में अभिव्यक्ति को योग और गुणनफल में बदलें।

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Question 12/30 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

यदि \(x^2-8x-20=0\) के मूल \(\alpha,\beta\) हैं, तो \(\alpha^2\beta+\alpha\beta^2\) क्या होगा?

If the roots of \(x^2-8x-20=0\) are \(\alpha,\beta\), what is \(\alpha^2\beta+\alpha\beta^2\)?

Explanation opens after your attempt
Correct Answer

A. (-160)

Step 1

Concept

(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=8\) and \(\alpha\beta=-20\), so the value is (-160). In exams, factor the expression first.

Step 2

Why this answer is correct

The correct answer is A. (-160). (\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=8\) and \(\alpha\beta=-20\), so the value is (-160). In exams, factor the expression first.

Step 3

Exam Tip

(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), जहां \(\alpha+\beta=8\) और \(\alpha\beta=-20\), इसलिए मान (-160) है। परीक्षा में अभिव्यक्ति को पहले factor करें।

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Question 13/30 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

यदि \(5x^2-17x+6=0\) के मूल \(\alpha,\beta\) हैं, तो (\(\alpha-\beta\)2) क्या होगा?

If \(\alpha,\beta\) are roots of \(5x^2-17x+6=0\), what is (\(\alpha-\beta\)2)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{169}{25}\)

Step 1

Concept

(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{17}{5}\right\)2-\frac{24}{5}=\frac{169}{25}). In exams, convert fractions to a common denominator.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{169}{25}\). (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{17}{5}\right\)2-\frac{24}{5}=\frac{169}{25}). In exams, convert fractions to a common denominator.

Step 3

Exam Tip

(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{17}{5}\right\)2-\frac{24}{5}=\frac{169}{25}) है। परीक्षा में भिन्नों को समान हर में बदलें।

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Question 14/30 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

यदि \(x^2-16x+63=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) क्या होगा?

If the roots of \(x^2-16x+63=0\) are \(\alpha\) and \(\beta\), what is \(\frac{1}{\alpha}+\frac{1}{\beta}\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{16}{63}\)

Step 1

Concept

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{16}{63}\). In exams, first write sum and product in reciprocal questions.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{16}{63}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{16}{63}\). In exams, first write sum and product in reciprocal questions.

Step 3

Exam Tip

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{16}{63}\) होता है। परीक्षा में reciprocal वाले प्रश्न में पहले योग और गुणनफल लिखें।

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Question 15/30 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

यदि \(x^2-19x+88=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो \(\alpha^2+\beta^2\) क्या होगा?

If the roots of \(x^2-19x+88=0\) are \(\alpha\) and \(\beta\), what is \(\alpha^2+\beta^2\)?

Explanation opens after your attempt
Correct Answer

A. (185)

Step 1

Concept

\(\alpha+\beta=19\) and \(\alpha\beta=88\), so (\alpha-2+\beta-2=192-2(88)=185). In exams, remember (\(\alpha+\beta\)2-2\alpha\beta).

Step 2

Why this answer is correct

The correct answer is A. (185). \(\alpha+\beta=19\) and \(\alpha\beta=88\), so (\alpha-2+\beta-2=192-2(88)=185). In exams, remember (\(\alpha+\beta\)2-2\alpha\beta).

Step 3

Exam Tip

\(\alpha+\beta=19\) और \(\alpha\beta=88\), इसलिए (\alpha-2+\beta-2=192-2(88)=185) है। परीक्षा में (\(\alpha+\beta\)2-2\alpha\beta) याद रखें।

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Question 16/30 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

यदि \(x^2-14x+45=0\) के मूल \(\alpha,\beta\) हैं, तो \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) क्या होगा?

If the roots of \(x^2-14x+45=0\) are \(\alpha,\beta\), what is \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{106}{45}\)

Step 1

Concept

\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=14\) and \(\alpha\beta=45\), so the value is \(\frac{196-90}{45}=\frac{106}{45}\). In exams, convert expressions into sum and product.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{106}{45}\). \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=14\) and \(\alpha\beta=45\), so the value is \(\frac{196-90}{45}=\frac{106}{45}\). In exams, convert expressions into sum and product.

Step 3

Exam Tip

\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), जहां \(\alpha+\beta=14\) और \(\alpha\beta=45\), इसलिए मान \(\frac{196-90}{45}=\frac{106}{45}\) है। परीक्षा में अभिव्यक्ति को योग और गुणनफल में बदलें।

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Question 17/30 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

यदि \(x^2-6x-16=0\) के मूल \(\alpha,\beta\) हैं, तो \(\alpha^2\beta+\alpha\beta^2\) क्या होगा?

If the roots of \(x^2-6x-16=0\) are \(\alpha,\beta\), what is \(\alpha^2\beta+\alpha\beta^2\)?

Explanation opens after your attempt
Correct Answer

A. (-96)

Step 1

Concept

(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=6\) and \(\alpha\beta=-16\), so the value is (-96). In exams, factor the expression first.

Step 2

Why this answer is correct

The correct answer is A. (-96). (\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=6\) and \(\alpha\beta=-16\), so the value is (-96). In exams, factor the expression first.

Step 3

Exam Tip

(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), जहां \(\alpha+\beta=6\) और \(\alpha\beta=-16\), इसलिए मान (-96) है। परीक्षा में अभिव्यक्ति को पहले factor करें।

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Question 18/30 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

यदि \(4x^2-13x+3=0\) के मूल \(\alpha,\beta\) हैं, तो (\(\alpha-\beta\)2) क्या होगा?

If \(\alpha,\beta\) are roots of \(4x^2-13x+3=0\), what is (\(\alpha-\beta\)2)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{121}{16}\)

Step 1

Concept

(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{13}{4}\right\)2-3=\frac{121}{16}). In exams, use this identity for the square of difference.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{121}{16}\). (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{13}{4}\right\)2-3=\frac{121}{16}). In exams, use this identity for the square of difference.

Step 3

Exam Tip

(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{13}{4}\right\)2-3=\frac{121}{16}) है। परीक्षा में अंतर का वर्ग इस पहचान से निकालें।

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Question 19/30 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

यदि \(x^2-12x+35=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) क्या होगा?

If the roots of \(x^2-12x+35=0\) are \(\alpha\) and \(\beta\), what is \(\frac{1}{\alpha}+\frac{1}{\beta}\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{12}{35}\)

Step 1

Concept

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{12}{35}\). In exams, first write sum and product in reciprocal questions.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{12}{35}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{12}{35}\). In exams, first write sum and product in reciprocal questions.

Step 3

Exam Tip

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{12}{35}\) होता है। परीक्षा में reciprocal वाले प्रश्न में पहले योग और गुणनफल लिखें।

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Question 20/30 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

यदि \(x^2-17x+70=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो \(\alpha^2+\beta^2\) क्या होगा?

If the roots of \(x^2-17x+70=0\) are \(\alpha\) and \(\beta\), what is \(\alpha^2+\beta^2\)?

Explanation opens after your attempt
Correct Answer

A. (149)

Step 1

Concept

\(\alpha+\beta=17\) and \(\alpha\beta=70\), so (\alpha-2+\beta-2=172-2(70)=149). In exams, remember (\(\alpha+\beta\)2-2\alpha\beta).

Step 2

Why this answer is correct

The correct answer is A. (149). \(\alpha+\beta=17\) and \(\alpha\beta=70\), so (\alpha-2+\beta-2=172-2(70)=149). In exams, remember (\(\alpha+\beta\)2-2\alpha\beta).

Step 3

Exam Tip

\(\alpha+\beta=17\) और \(\alpha\beta=70\), इसलिए (\alpha-2+\beta-2=172-2(70)=149) है। परीक्षा में (\(\alpha+\beta\)2-2\alpha\beta) याद रखें।

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Question 21/30 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

यदि \(x^2-10x+21=0\) के मूल \(\alpha,\beta\) हैं, तो \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) क्या होगा?

If the roots of \(x^2-10x+21=0\) are \(\alpha,\beta\), what is \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{58}{21}\)

Step 1

Concept

\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=10\) and \(\alpha\beta=21\), so the value is \(\frac{100-42}{21}=\frac{58}{21}\). In exams, convert expressions into sum and product.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{58}{21}\). \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=10\) and \(\alpha\beta=21\), so the value is \(\frac{100-42}{21}=\frac{58}{21}\). In exams, convert expressions into sum and product.

Step 3

Exam Tip

\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), जहां \(\alpha+\beta=10\) और \(\alpha\beta=21\), इसलिए मान \(\frac{100-42}{21}=\frac{58}{21}\) है। परीक्षा में अभिव्यक्ति को योग और गुणनफल में बदलें।

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Question 22/30 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

यदि \(x^2-4x-12=0\) के मूल \(\alpha,\beta\) हैं, तो \(\alpha^2\beta+\alpha\beta^2\) क्या होगा?

If the roots of \(x^2-4x-12=0\) are \(\alpha,\beta\), what is \(\alpha^2\beta+\alpha\beta^2\)?

Explanation opens after your attempt
Correct Answer

A. (-48)

Step 1

Concept

(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=4\) and \(\alpha\beta=-12\), so the value is (-48). In exams, factor the expression first.

Step 2

Why this answer is correct

The correct answer is A. (-48). (\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=4\) and \(\alpha\beta=-12\), so the value is (-48). In exams, factor the expression first.

Step 3

Exam Tip

(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), जहां \(\alpha+\beta=4\) और \(\alpha\beta=-12\), इसलिए मान (-48) है। परीक्षा में अभिव्यक्ति को पहले factor करें।

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Question 23/30 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

यदि \(3x^2-11x+6=0\) के मूल \(\alpha,\beta\) हैं, तो (\(\alpha-\beta\)2) क्या होगा?

If \(\alpha,\beta\) are roots of \(3x^2-11x+6=0\), what is (\(\alpha-\beta\)2)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{49}{9}\)

Step 1

Concept

(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{11}{3}\right\)2-8=\frac{49}{9}). In exams, use this identity for the square of difference.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{49}{9}\). (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{11}{3}\right\)2-8=\frac{49}{9}). In exams, use this identity for the square of difference.

Step 3

Exam Tip

(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{11}{3}\right\)2-8=\frac{49}{9}) है। परीक्षा में अंतर का वर्ग इस पहचान से निकालें।

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Question 24/30 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

यदि \(x^2-9x+20=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) क्या होगा?

If the roots of \(x^2-9x+20=0\) are \(\alpha\) and \(\beta\), what is \(\frac{1}{\alpha}+\frac{1}{\beta}\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{9}{20}\)

Step 1

Concept

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{9}{20}\). In exams, first write sum and product in reciprocal questions.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{9}{20}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{9}{20}\). In exams, first write sum and product in reciprocal questions.

Step 3

Exam Tip

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{9}{20}\) होता है। परीक्षा में reciprocal वाले प्रश्न में पहले योग और गुणनफल लिखें।

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Question 25/30 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

यदि \(x^2-13x+40=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो \(\alpha^2+\beta^2\) क्या होगा?

If the roots of \(x^2-13x+40=0\) are \(\alpha\) and \(\beta\), what is \(\alpha^2+\beta^2\)?

Explanation opens after your attempt
Correct Answer

A. (89)

Step 1

Concept

\(\alpha+\beta=13\) and \(\alpha\beta=40\), so (\alpha-2+\beta-2=132-2(40)=89). In exams, remember (\(\alpha+\beta\)2-2\alpha\beta).

Step 2

Why this answer is correct

The correct answer is A. (89). \(\alpha+\beta=13\) and \(\alpha\beta=40\), so (\alpha-2+\beta-2=132-2(40)=89). In exams, remember (\(\alpha+\beta\)2-2\alpha\beta).

Step 3

Exam Tip

\(\alpha+\beta=13\) और \(\alpha\beta=40\), इसलिए (\alpha-2+\beta-2=132-2(40)=89) है। परीक्षा में (\(\alpha+\beta\)2-2\alpha\beta) याद रखें।

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Question 26/30 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

यदि \(x^2-8x+15=0\) के मूल \(\alpha,\beta\) हैं, तो \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) क्या होगा?

If the roots of \(x^2-8x+15=0\) are \(\alpha,\beta\), what is \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{34}{15}\)

Step 1

Concept

\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=8\) and \(\alpha\beta=15\), so the value is \(\frac{64-30}{15}=\frac{34}{15}\). In exams, convert expressions into sum and product.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{34}{15}\). \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=8\) and \(\alpha\beta=15\), so the value is \(\frac{64-30}{15}=\frac{34}{15}\). In exams, convert expressions into sum and product.

Step 3

Exam Tip

\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), जहां \(\alpha+\beta=8\) और \(\alpha\beta=15\), इसलिए मान \(\frac{64-30}{15}=\frac{34}{15}\) है। परीक्षा में अभिव्यक्ति को योग और गुणनफल में बदलें।

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Question 27/30 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

यदि \(x^2-2x-8=0\) के मूल \(\alpha,\beta\) हैं, तो \(\alpha^2\beta+\alpha\beta^2\) क्या होगा?

If the roots of \(x^2-2x-8=0\) are \(\alpha,\beta\), what is \(\alpha^2\beta+\alpha\beta^2\)?

Explanation opens after your attempt
Correct Answer

A. (-16)

Step 1

Concept

(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=2\) and \(\alpha\beta=-8\), so the value is (-16). In exams, factor the expression first.

Step 2

Why this answer is correct

The correct answer is A. (-16). (\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=2\) and \(\alpha\beta=-8\), so the value is (-16). In exams, factor the expression first.

Step 3

Exam Tip

(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), जहां \(\alpha+\beta=2\) और \(\alpha\beta=-8\), इसलिए मान (-16) है। परीक्षा में अभिव्यक्ति को पहले factor करें।

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Question 28/30 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

यदि \(2x^2-9x+4=0\) के मूल \(\alpha,\beta\) हैं, तो (\(\alpha-\beta\)2) क्या होगा?

If \(\alpha,\beta\) are roots of \(2x^2-9x+4=0\), what is (\(\alpha-\beta\)2)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{65}{4}\)

Step 1

Concept

(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{9}{2}\right\)2-8=\frac{65}{4}). In exams, use this identity for square of difference.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{65}{4}\). (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{9}{2}\right\)2-8=\frac{65}{4}). In exams, use this identity for square of difference.

Step 3

Exam Tip

(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{9}{2}\right\)2-8=\frac{65}{4}) है। परीक्षा में अंतर का वर्ग सीधे इस पहचान से निकालें।

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Question 29/30 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

यदि \(x^2-7x+10=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) क्या होगा?

If the roots of \(x^2-7x+10=0\) are \(\alpha\) and \(\beta\), what is \(\frac{1}{\alpha}+\frac{1}{\beta}\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{7}{10}\)

Step 1

Concept

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{7}{10}\). In exams, first write sum and product for reciprocal questions.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{7}{10}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{7}{10}\). In exams, first write sum and product for reciprocal questions.

Step 3

Exam Tip

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{7}{10}\) होता है। परीक्षा में reciprocal वाले प्रश्न में पहले योग और गुणनफल लिखें।

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Question 30/30 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

यदि \(x^2-11x+30=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो \(\alpha^2+\beta^2\) क्या होगा?

If the roots of \(x^2-11x+30=0\) are \(\alpha\) and \(\beta\), what is \(\alpha^2+\beta^2\)?

Explanation opens after your attempt
Correct Answer

A. (61)

Step 1

Concept

\(\alpha+\beta=11\) and \(\alpha\beta=30\), so (\alpha-2+\beta-2=(11)2-2(30)=61). In exams, remember the identity (\(\alpha+\beta\)2-2\alpha\beta).

Step 2

Why this answer is correct

The correct answer is A. (61). \(\alpha+\beta=11\) and \(\alpha\beta=30\), so (\alpha-2+\beta-2=(11)2-2(30)=61). In exams, remember the identity (\(\alpha+\beta\)2-2\alpha\beta).

Step 3

Exam Tip

\(\alpha+\beta=11\) और \(\alpha\beta=30\), इसलिए (\alpha-2+\beta-2=(11)2-2(30)=61) है। परीक्षा में पहचान (\(\alpha+\beta\)2-2\alpha\beta) याद रखें।

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