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2 results found for "product of radicals" in Class 10.

Question Expert Mathematics Chapter 1: Real Numbers 5: Irrational numbers Class 10 Level 15

कौन सा कथन \(\sqrt{2}\cdot \sqrt{3}\) के बारे में सही है?

Which statement is correct about \(\sqrt{2}\cdot \sqrt{3}\)?

Explanation opens after your attempt
Correct Answer

B. यह \(\sqrt{6}\) है और अपरिमेय हैIt is \(\sqrt{6}\) and irrational

Step 1

Concept

The product of radicals is \(\sqrt{2}\cdot \sqrt{3}=\sqrt{6}\).

Step 2

Why this answer is correct

Since (6) is not a perfect square \(\sqrt{6}\) is irrational.

Step 3

Exam Tip

In multiplication the numbers inside radicals multiply, not add. चरण 1: वर्गमूलों का गुणनफल \(\sqrt{2}\cdot \sqrt{3}=\sqrt{6}\) है। चरण 2: (6) पूर्ण वर्ग नहीं है इसलिए \(\sqrt{6}\) अपरिमेय है। चरण 3: गुणन में भीतर की संख्याएं गुणा होती हैं जोड़ नहीं।

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Question Expert Mathematics Chapter 1: Real Numbers 5: Irrational numbers Class 10 Level 14

कौन-सा विकल्प \(\sqrt{a}\times\sqrt{b}\) को अपरिमेय बनाता है?

Which option makes \(\sqrt{a}\times\sqrt{b}\) irrational?

Explanation opens after your attempt
Correct Answer

D. (a=6,b=15)

Step 1

Concept

\(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\).

Step 2

Why this answer is correct

For (a=6,b=15), (ab=90), which is not a perfect square, so \(\sqrt{90}\) is irrational.

Step 3

Exam Tip

In multiplication, the key check is whether the product inside the root is a perfect square. चरण 1: \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\) होता है। चरण 2: (a=6,b=15) पर (ab=90), जो पूर्ण वर्ग नहीं है, इसलिए \(\sqrt{90}\) अपरिमेय है। चरण 3: गुणन में अंदर का गुणनफल पूर्ण वर्ग है या नहीं, यह मुख्य जाँच है।

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