NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 2: Polynomials - Exercise 2.3
Question 6

Question. 6

Find the zeroes of \(4x^2+5\sqrt{2}x-3\) and verify relations.

Answer:

\(x=\dfrac{\sqrt{2}}{4},\; x=-\dfrac{3\sqrt{2}}{2}\)

Detailed Answer with Explanation:

Step 1: Quadratic formula.

\(a = 4, b = 5\sqrt{2}, c = -3\)

Discriminant = \((5\sqrt{2})^2 - 4(4)(-3)\)

= 50 + 48 = 98

\(\sqrt{98} = 7\sqrt{2}\)

Step 2: Roots.

\(x = \dfrac{-5\sqrt{2} + 7\sqrt{2}}{8} = \dfrac{\sqrt{2}}{4}\)

\(x = \dfrac{-5\sqrt{2} - 7\sqrt{2}}{8} = -\dfrac{3\sqrt{2}}{2}\)

Step 3: Verify.

Sum = \(\dfrac{\sqrt{2}}{4} - \dfrac{3\sqrt{2}}{2} = -5\sqrt{2}/4\)

= \(-b/a = -(5\sqrt{2})/4\)

Product = \((\dfrac{\sqrt{2}}{4})(-\dfrac{3\sqrt{2}}{2}) = -3/4\)

= \(c/a = -3/4\)

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