NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 2: Polynomials - Exercise 2.3
Question 9

Question. 9

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

Answer:

\(y=\dfrac{-\sqrt{3}+\sqrt{83}}{4},\; y=\dfrac{-\sqrt{3}-\sqrt{83}}{4}\)

Detailed Answer with Explanation:

Step 1: Identify coefficients.

\(a=1, b=\sqrt{3}/2, c=-5\)

Step 2: Discriminant.

\(b^2 - 4ac = (\sqrt{3}/2)^2 - 4(1)(-5)\)

= 3/4 + 20 = 83/4

\(\sqrt{83/4} = \sqrt{83}/2\)

Step 3: Roots.

\(y = \dfrac{-\sqrt{3}/2 ± \sqrt{83}/2}{2}\)

= \(\dfrac{-\sqrt{3} ± \sqrt{83}}{4}\)

Step 4: Verify.

Sum = \((-\sqrt{3} + \sqrt{83})/4 + (-\sqrt{3} - \sqrt{83})/4\)

= -2\sqrt{3}/4 = -\sqrt{3}/2 = -b/a

Product = \((-\sqrt{3} + \sqrt{83})(-\sqrt{3} - \sqrt{83})/16\)

= (83 - 3)/16 = 80/16 = 5 = c/a

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