Concept-wise Practice

irrational roots MCQ Questions for Class 10

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Practice Questions

70 questions tagged with irrational roots.

यदि (p(x)=x-2-4x-1), तो इसके शून्यक कौन से हैं?

If (p(x)=x-2-4x-1), what are its zeroes?

Explanation opens after your attempt
Correct Answer

A. \(2+\sqrt{5}\) और \(2-\sqrt{5}\)\(2+\sqrt{5}\) and \(2-\sqrt{5}\)

Step 1

Concept

Using the quadratic formula, \(x=\frac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt{5}\). In exams simplify the discriminant.

Step 2

Why this answer is correct

The correct answer is A. \(2+\sqrt{5}\) और \(2-\sqrt{5}\) / \(2+\sqrt{5}\) and \(2-\sqrt{5}\). Using the quadratic formula, \(x=\frac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt{5}\). In exams simplify the discriminant.

Step 3

Exam Tip

द्विघात सूत्र से \(x=\frac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt{5}\) मिलता है। परीक्षा में विविक्तकर को सरल करें।

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किस मान के लिए \(x^2-2kx+2=0\) के मूल अपरिमेय और वास्तविक होंगे?

For which value of (k) will the roots of \(x^2-2kx+2=0\) be irrational and real?

Explanation opens after your attempt
Correct Answer

B. (k=2)

Step 1

Concept

For (k=2), the discriminant is (16-8=8), positive but not a perfect square. Therefore the roots are real and irrational.

Step 2

Why this answer is correct

The correct answer is B. (k=2). For (k=2), the discriminant is (16-8=8), positive but not a perfect square. Therefore the roots are real and irrational.

Step 3

Exam Tip

(k=2) पर विविक्तकर (16-8=8), जो धनात्मक पर पूर्ण वर्ग नहीं है। इसलिए मूल वास्तविक और अपरिमेय होंगे।

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यदि \(x^2-6x+1=0\), तो (x) के मान किस प्रकार के हैं?

If \(x^2-6x+1=0\), what type are the values of (x)?

Explanation opens after your attempt
Correct Answer

C. दो अलग अपरिमेय वास्तविकTwo distinct irrational real values

Step 1

Concept

The discriminant is (36-4=32), positive but not a perfect square. So there are two distinct irrational real roots.

Step 2

Why this answer is correct

The correct answer is C. दो अलग अपरिमेय वास्तविक / Two distinct irrational real values. The discriminant is (36-4=32), positive but not a perfect square. So there are two distinct irrational real roots.

Step 3

Exam Tip

विविक्तकर (36-4=32) है, जो पूर्ण वर्ग नहीं है और धनात्मक है। इसलिए दो अलग अपरिमेय वास्तविक मूल मिलते हैं।

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किस बहुपद के शून्यक \(3+\sqrt{2}\) और \(3-\sqrt{2}\) हैं?

Which polynomial has zeroes \(3+\sqrt{2}\) and \(3-\sqrt{2}\)?

Explanation opens after your attempt
Correct Answer

A. \(x^2-6x+7\)

Step 1

Concept

The sum is (6) and the product is (9-2=7). The polynomial is \(x^2-6x+7\).

Step 2

Why this answer is correct

The correct answer is A. \(x^2-6x+7\). The sum is (6) and the product is (9-2=7). The polynomial is \(x^2-6x+7\).

Step 3

Exam Tip

योग (6) और गुणनफल (9-2=7) है। बहुपद \(x^2-6x+7\) होगा।

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यदि (p(x)=x-2-k) के शून्यक अपरिमेय वास्तविक हैं, तो (k) के लिए सही शर्त कौन सी है?

If the zeroes of (p(x)=x-2-k) are irrational real, which condition on (k) is correct?

Explanation opens after your attempt
Correct Answer

B. (k) धनात्मक हो लेकिन पूर्ण वर्ग न हो(k) is positive but not a perfect square

Step 1

Concept

The zeroes are \(x=\pm\sqrt{k}\). They are irrational real when (k>0) and (k) is not a perfect square.

Step 2

Why this answer is correct

The correct answer is B. (k) धनात्मक हो लेकिन पूर्ण वर्ग न हो / (k) is positive but not a perfect square. The zeroes are \(x=\pm\sqrt{k}\). They are irrational real when (k>0) and (k) is not a perfect square.

Step 3

Exam Tip

शून्यक \(x=\pm\sqrt{k}\) हैं। ये अपरिमेय वास्तविक तभी होंगे जब (k>0) और (k) पूर्ण वर्ग न हो।

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यदि (p(x)=x-2-2) है, तो (p(x)) के शून्यकों के बारे में सही कथन कौन सा है?

If (p(x)=x-2-2), which statement about the zeroes of (p(x)) is correct?

Explanation opens after your attempt
Correct Answer

B. दोनों अपरिमेय वास्तविक हैंBoth are irrational real

Step 1

Concept

From \(x^2-2=0\), \(x=\pm\sqrt{2}\), which are irrational real numbers. In exams, check both real nature and rationality of roots.

Step 2

Why this answer is correct

The correct answer is B. दोनों अपरिमेय वास्तविक हैं / Both are irrational real. From \(x^2-2=0\), \(x=\pm\sqrt{2}\), which are irrational real numbers. In exams, check both real nature and rationality of roots.

Step 3

Exam Tip

\(x^2-2=0\) से \(x=\pm\sqrt{2}\), जो अपरिमेय वास्तविक संख्याएँ हैं। परीक्षा में मूल निकालते समय वास्तविकता और परिमेयता दोनों जाँचें।

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यदि (p(x)=x-2+ax+6) का एक शून्यक \(-\sqrt{6}\) है और दूसरा \(\sqrt{6}\) है, तो (a) क्या है?

If one zero of (p(x)=x-2+ax+6) is \(-\sqrt{6}\) and the other is \(\sqrt{6}\), what is (a)?

Explanation opens after your attempt
Correct Answer

A. (0)

Step 1

Concept

The sum of the zeroes is (0), and in \(x^2+ax+6\), the sum is (-a). Thus (a=0).

Step 2

Why this answer is correct

The correct answer is A. (0). The sum of the zeroes is (0), and in \(x^2+ax+6\), the sum is (-a). Thus (a=0).

Step 3

Exam Tip

शून्यकों का योग (0) है और \(x^2+ax+6\) में योग (-a) होता है। अतः (a=0)।

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यदि किसी द्विघात बहुपद के परिमेय गुणांक हैं और एक शून्यक \(3+\sqrt{5}\) है, तो दूसरा शून्यक क्या होगा?

If a quadratic polynomial has rational coefficients and one zero is \(3+\sqrt{5}\), what will be the other zero?

Explanation opens after your attempt
Correct Answer

A. \(3-\sqrt{5}\)

Step 1

Concept

For a quadratic with rational coefficients, \(a-\sqrt{b}\) accompanies \(a+\sqrt{b}\). Remember this as the conjugate-zero rule.

Step 2

Why this answer is correct

The correct answer is A. \(3-\sqrt{5}\). For a quadratic with rational coefficients, \(a-\sqrt{b}\) accompanies \(a+\sqrt{b}\). Remember this as the conjugate-zero rule.

Step 3

Exam Tip

परिमेय गुणांकों वाले द्विघात में \(a+\sqrt{b}\) के साथ \(a-\sqrt{b}\) भी शून्यक होता है। परीक्षा में इसे संयुग्मी शून्यक नियम की तरह याद रखें।

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\(\sqrt{2}\), \(\sqrt{3}\) और \(\sqrt{5}\) को पूर्णांक क्यों नहीं माना जाता?

Why are \(\sqrt{2}\), \(\sqrt{3}\), and \(\sqrt{5}\) not taken as integers?

Explanation opens after your attempt
Correct Answer

A. क्योंकि (2), (3), और (5) पूर्ण वर्ग नहीं हैंBecause (2), (3), and (5) are not perfect squares

Step 1

Concept

The square root of a perfect square is an integer.

Step 2

Why this answer is correct

(2), (3), and (5) are not perfect squares.

Step 3

Exam Tip

Therefore their square roots are proved irrational. चरण 1: किसी पूर्ण वर्ग का वर्गमूल पूर्णांक होता है। चरण 2: (2), (3), और (5) पूर्ण वर्ग नहीं हैं। चरण 3: इसलिए इनके वर्गमूलों की अपरिमेयता सिद्ध की जाती है।

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किस कारण से \(\sqrt{2}\), \(\sqrt{3}\) और \(\sqrt{5}\) को सीधे पूर्णांक नहीं माना जा सकता?

Why can \(\sqrt{2}\), \(\sqrt{3}\), and \(\sqrt{5}\) not be directly treated as integers?

Explanation opens after your attempt
Correct Answer

A. क्योंकि (2), (3), और (5) पूर्ण वर्ग नहीं हैंBecause (2), (3), and (5) are not perfect squares

Step 1

Concept

Square roots of perfect squares are integers.

Step 2

Why this answer is correct

(2), (3), and (5) are not perfect squares.

Step 3

Exam Tip

That is why irrationality proofs are studied for their square roots. चरण 1: पूर्ण वर्गों के वर्गमूल पूर्णांक होते हैं। चरण 2: (2), (3), और (5) पूर्ण वर्ग नहीं हैं। चरण 3: इसलिए इनके वर्गमूलों के लिए अपरिमेयता का प्रमाण पढ़ाया जाता है।

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