NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 2: Polynomials - Exercise 2.3
Question 7

Question. 7

Find the zeroes of \(2s^2-(1+2\sqrt{2})s+\sqrt{2}\) and verify relations.

Answer:

\(s=\dfrac{1}{2},\; s=\sqrt{2}\)

Detailed Answer with Explanation:

Step 1: Check zeroes.

Substitute \(s = \sqrt{2}\)

\(2(2) - (1 + 2\sqrt{2})(\sqrt{2}) + \sqrt{2}\)

= 4 - \sqrt{2} - 4 + \sqrt{2} = 0

So, \(s = \sqrt{2}\) is a root.

Substitute \(s = 1/2\)

\(2(1/4) - (1 + 2\sqrt{2})(1/2) + \sqrt{2}\)

= 1/2 - 1/2 - \sqrt{2} + \sqrt{2} = 0

So, \(s = 1/2\) is a root.

Step 2: Verify.

Sum = \(1/2 + \sqrt{2}\)

= \((1 + 2\sqrt{2})/2 = -b/a\)

Product = \((1/2)(\sqrt{2}) = \sqrt{2}/2\)

= c/a

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 2: Polynomials – Exercise 2.3 | Detailed Answers