Search Class 10 Questions

15 results found for "irrational root" in Class 10.

Question Hard Mathematics Polynomials Irrational numbers and real numbers Class 10 Level 26

यदि \(x=1-\sqrt{5}\), तो \(x^2-2x-4\) का मान क्या है?

If \(x=1-\sqrt{5}\), what is the value of \(x^2-2x-4\)?

Explanation opens after your attempt
Correct Answer

A. (0)

Step 1

Concept

Since \(x-1=-\sqrt{5}\), ((x-1)2=5), so \(x^2-2x-4=0\). Isolate the irrational part and square in exams.

Step 2

Why this answer is correct

The correct answer is A. (0). Since \(x-1=-\sqrt{5}\), ((x-1)2=5), so \(x^2-2x-4=0\). Isolate the irrational part and square in exams.

Step 3

Exam Tip

\(x-1=-\sqrt{5}\), इसलिए ((x-1)2=5) से \(x^2-2x-4=0\) मिलता है। परीक्षा में अपरिमेय भाग अलग करके वर्ग करें।

Open Question Page
Ask Friends
Question Hard Mathematics Polynomials Irrational numbers and real numbers Class 10 Level 26

यदि \(x=\sqrt{2}\) बहुपद \(ax^2+bx+c\) का शून्यक है और (a,b,c) परिमेय हैं, तो कौन सा निष्कर्ष सही नहीं हो सकता?

If \(x=\sqrt{2}\) is a zero of \(ax^2+bx+c\) and (a,b,c) are rational, which conclusion cannot be correct?

Explanation opens after your attempt
Correct Answer

C. सिर्फ \(\sqrt{2}\) ही अकेला अपरिमेय शून्यक हो और गुणांक परिमेय रहेंOnly \(\sqrt{2}\) is the sole irrational zero while coefficients stay rational

Step 1

Concept

In a quadratic with rational coefficients an irrational zero comes with its conjugate. In exams be suspicious of a lone irrational root.

Step 2

Why this answer is correct

The correct answer is C. सिर्फ \(\sqrt{2}\) ही अकेला अपरिमेय शून्यक हो और गुणांक परिमेय रहें / Only \(\sqrt{2}\) is the sole irrational zero while coefficients stay rational. In a quadratic with rational coefficients an irrational zero comes with its conjugate. In exams be suspicious of a lone irrational root.

Step 3

Exam Tip

परिमेय गुणांकों वाले द्विघात में अपरिमेय शून्यक अपने संयुग्मी के साथ आता है। परीक्षा में अकेले अपरिमेय मूल पर संदेह करें।

Open Question Page
Ask Friends
Question Hard Mathematics Polynomials Irrational numbers and real numbers Class 10 Level 26

यदि \(x=3-\sqrt{2}\), तो \(x^2-6x+7\) का मान क्या है?

If \(x=3-\sqrt{2}\), what is the value of \(x^2-6x+7\)?

Explanation opens after your attempt
Correct Answer

A. (0)

Step 1

Concept

Since \(x-3=-\sqrt{2}\), ((x-3)2=2), hence \(x^2-6x+7=0\). Handle \(a-\sqrt{b}\) the same way in exams.

Step 2

Why this answer is correct

The correct answer is A. (0). Since \(x-3=-\sqrt{2}\), ((x-3)2=2), hence \(x^2-6x+7=0\). Handle \(a-\sqrt{b}\) the same way in exams.

Step 3

Exam Tip

\(x-3=-\sqrt{2}\), इसलिए ((x-3)2=2) से \(x^2-6x+7=0\) है। परीक्षा में \(a-\sqrt{b}\) को भी इसी विधि से संभालें।

Open Question Page
Ask Friends
Question Hard Mathematics Polynomials Irrational numbers and real numbers Class 10 Level 26

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।

Open Question Page
Ask Friends
Question Hard Mathematics Polynomials Irrational numbers and real numbers Class 10 Level 28

यदि \(x^2+2x+5=0\), तो वास्तविक संख्या पद्धति में मूलों के बारे में कौन सा कथन सही है?

If \(x^2+2x+5=0\), which statement about the roots in the real number system is correct?

Explanation opens after your attempt
Correct Answer

C. कोई वास्तविक मूल नहीं हैThere are no real roots

Step 1

Concept

The discriminant is (4-20=-16), which is negative, so there are no real roots. In exams do not treat a negative discriminant as real zeroes.

Step 2

Why this answer is correct

The correct answer is C. कोई वास्तविक मूल नहीं है / There are no real roots. The discriminant is (4-20=-16), which is negative, so there are no real roots. In exams do not treat a negative discriminant as real zeroes.

Step 3

Exam Tip

विविक्तकर (4-20=-16) ऋणात्मक है, इसलिए वास्तविक मूल नहीं हैं। परीक्षा में ऋणात्मक विविक्तकर को वास्तविक शून्यक नहीं मानें।

Open Question Page
Ask Friends
Question Hard Mathematics Polynomials Irrational numbers and real numbers Class 10 Level 28

यदि \(x^2-4x+4=0\), तो मूलों की प्रकृति क्या है?

If \(x^2-4x+4=0\), what is the nature of the roots?

Explanation opens after your attempt
Correct Answer

B. दो समान परिमेयTwo equal rational roots

Step 1

Concept

The discriminant is (16-16=0), so the roots are equal and (x=2) is rational. In exams (D=0) means equal roots.

Step 2

Why this answer is correct

The correct answer is B. दो समान परिमेय / Two equal rational roots. The discriminant is (16-16=0), so the roots are equal and (x=2) is rational. In exams (D=0) means equal roots.

Step 3

Exam Tip

विविक्तकर (16-16=0), इसलिए मूल समान हैं और (x=2) परिमेय है। परीक्षा में (D=0) का अर्थ समान मूल होता है।

Open Question Page
Ask Friends
Question Hard Mathematics Polynomials Irrational numbers and real numbers Class 10 Level 28

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।

Open Question Page
Ask Friends
Question Expert Mathematics Polynomials Irrational numbers and real numbers Class 10 Level 25

यदि (p(x)=x-2-16) है, तो शून्यकों के प्रकार के बारे में सही कथन कौन सा है?

If (p(x)=x-2-16), which statement about the type of zeroes is correct?

Explanation opens after your attempt
Correct Answer

A. दोनों परिमेय वास्तविक हैंBoth are rational real

Step 1

Concept

From \(x^2-16=0\), \(x=\pm4\), which are rational real. Not every square-root type question gives irrational roots.

Step 2

Why this answer is correct

The correct answer is A. दोनों परिमेय वास्तविक हैं / Both are rational real. From \(x^2-16=0\), \(x=\pm4\), which are rational real. Not every square-root type question gives irrational roots.

Step 3

Exam Tip

\(x^2-16=0\) से \(x=\pm4\), जो परिमेय वास्तविक हैं। हर वर्गमूल वाला प्रश्न अपरिमेय नहीं होता।

Open Question Page
Ask Friends
Question Medium Mathematics Polynomials Irrational numbers and real numbers Class 10 Level 27

कौन सा कथन \(\sqrt{a}\) के लिए सही है जब (a) धनात्मक पूर्णांक है?

Which statement is correct for \(\sqrt{a}\) when (a) is a positive integer?

Explanation opens after your attempt
Correct Answer

A. यदि (a) पूर्ण वर्ग नहीं है तो \(\sqrt{a}\) अपरिमेय हैIf (a) is not a perfect square then \(\sqrt{a}\) is irrational

Step 1

Concept

The square root of a positive integer is rational only when the number is a perfect square. So a non perfect square gives an irrational root.

Step 2

Why this answer is correct

The correct answer is A. यदि (a) पूर्ण वर्ग नहीं है तो \(\sqrt{a}\) अपरिमेय है / If (a) is not a perfect square then \(\sqrt{a}\) is irrational. The square root of a positive integer is rational only when the number is a perfect square. So a non perfect square gives an irrational root.

Step 3

Exam Tip

धनात्मक पूर्णांक की वर्गमूल परिमेय तभी होती है जब संख्या पूर्ण वर्ग हो। इसलिए पूर्ण वर्ग न हो तो जड़ अपरिमेय होगी।

Open Question Page
Ask Friends
Question Easy Mathematics Polynomials Irrational numbers and real numbers Class 10 Level 26

कौन सा विकल्प केवल परिमेय संख्याएँ दिखाता है?

Which option shows only rational numbers?

Explanation opens after your attempt
Correct Answer

A. (4), (-2), (0.75)

Step 1

Concept

Integers and terminating decimals are rational. Options containing irrational roots or \(\pi\) are not only rational.

Step 2

Why this answer is correct

The correct answer is A. (4), (-2), (0.75). Integers and terminating decimals are rational. Options containing irrational roots or \(\pi\) are not only rational.

Step 3

Exam Tip

पूर्णांक और सांत दशमलव परिमेय होते हैं। जिन विकल्पों में अपरिमेय जड़ या \(\pi\) है वे केवल परिमेय नहीं हैं।

Open Question Page
Ask Friends
Question Easy Mathematics Real Numbers 6: Proof of irrationality of √2, √3, √5 Class 10 Level 18

\(\sqrt{2}\), \(\sqrt{3}\) 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।

Open Question Page
Ask Friends
Question Easy Mathematics Real Numbers 6: Proof of irrationality of √2, √3, √5 Class 10 Level 17

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।

Open Question Page
Ask Friends
Question Expert Mathematics Real Numbers 5: Irrational numbers Class 10 Level 13

यदि \(x=\sqrt{a}\) अपरिमेय है और (a<50) धनात्मक पूर्णांक है, तो कौन-सा (a) उपयुक्त नहीं है?

If \(x=\sqrt{a}\) is irrational and (a<50) is a positive integer, which (a) is not suitable?

Explanation opens after your attempt
Correct Answer

C. (36)

Step 1

Concept

\(\sqrt{a}\) is irrational only when (a) is not a perfect square.

Step 2

Why this answer is correct

(36) is a perfect square and \(\sqrt{36}=6\), so it is not suitable.

Step 3

Exam Tip

Quickly identify perfect squares among the options. चरण 1: \(\sqrt{a}\) अपरिमेय तभी होगा जब (a) पूर्ण वर्ग न हो। चरण 2: (36) पूर्ण वर्ग है और \(\sqrt{36}=6\), इसलिए यह उपयुक्त नहीं है। चरण 3: विकल्पों में पूर्ण वर्ग को तुरंत पहचानें।

Open Question Page
Ask Friends
Question Hard Mathematics Real Numbers 5: Irrational numbers Class 10 Level 15

किस विकल्प में \(\sqrt{a}\) अपरिमेय है?

In which option is \(\sqrt{a}\) irrational?

Explanation opens after your attempt
Correct Answer

C. (a=90)

Step 1

Concept

(49), (81), and (121) are perfect squares.

Step 2

Why this answer is correct

(90) is not a perfect square, so \(\sqrt{90}\) is irrational.

Step 3

Exam Tip

To decide the nature of a square root, first check perfect squares. चरण 1: (49), (81) और (121) पूर्ण वर्ग हैं। चरण 2: (90) पूर्ण वर्ग नहीं है, इसलिए \(\sqrt{90}\) अपरिमेय है। चरण 3: वर्गमूल की प्रकृति जानने के लिए सबसे पहले पूर्ण वर्ग जाँचें।

Open Question Page
Ask Friends
Question Hard Mathematics Real Numbers 5: Irrational numbers Class 10 Level 14

यदि (x) अपरिमेय है और \(x^2=7\), तो (x) का संभावित मान कौन-सा है?

If (x) is irrational and \(x^2=7\), which can be the value of (x)?

Explanation opens after your attempt
Correct Answer

B. \(\sqrt{7}\)

Step 1

Concept

From \(x^2=7\), \(x=\sqrt{7}\) or \(x=-\sqrt{7}\).

Step 2

Why this answer is correct

Among the options, \(\sqrt{7}\) is present and it is irrational.

Step 3

Exam Tip

Remember both positive and negative roots, then match the given options. चरण 1: \(x^2=7\) से \(x=\sqrt{7}\) या \(x=-\sqrt{7}\) हो सकता है। चरण 2: दिए गए विकल्पों में \(\sqrt{7}\) है, जो अपरिमेय है। चरण 3: वर्ग समीकरण में धन और ऋण दोनों मूल याद रखें, पर विकल्प के अनुसार चुनें।

Open Question Page
Ask Friends
Student Class Required

Select your class first

Quiz questions, daily challenge and practice pages will open according to your selected class. Class 11/12 ke liye stream bhi select karein.