NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 4: Quadatric Equation - Exercise 4.3
Question 1

Question. 1

Find the roots of the quadratic equations by using the quadratic formula in each of the following:

(i) \(2x^2 - 3x - 5 = 0\)

(ii) \(5x^2 + 13x + 8 = 0\)

(iii) \(-3x^2 + 5x + 12 = 0\)

(iv) \(-x^2 + 7x - 10 = 0\)

(v) \(x^2 + 2\sqrt{2}\,x - 6 = 0\)

(vi) \(x^2 - 3\sqrt{5}\,x + 10 = 0\)

(vii) \(\dfrac{1}{2}x^2 - \sqrt{11}\,x + 1 = 0\)

Answer:

(i) \(x=\dfrac{5}{2}\), \(x=-1\)

(ii) \(x=-1\), \(x=-\dfrac{8}{5}\)

(iii) \(x=3\), \(x=-\dfrac{4}{3}\)

(iv) \(x=5\), \(x=2\)

(v) \(x=\sqrt{2}\), \(x=-3\sqrt{2}\)

(vi) \(x=2\sqrt{5}\), \(x=\sqrt{5}\)

(vii) \(x=\sqrt{11}+3\), \(x=\sqrt{11}-3\)

Detailed Answer with Explanation:

We will solve all problems using the quadratic formula:

\[ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(a\), \(b\), and \(c\) are the coefficients of \(x^2\), \(x\), and the constant term in the equation.

\(D = b^2 - 4ac\) is called the discriminant.

(i) \(2x^2 - 3x - 5 = 0\)

Identify: \(a = 2\), \(b = -3\), \(c = -5\).

Discriminant: \(D = (-3)^2 - 4(2)(-5) = 9 + 40 = 49\).

Formula: \(x = \dfrac{-(-3) \pm \sqrt{49}}{2(2)} = \dfrac{3 \pm 7}{4}\).

So, \(x = \dfrac{10}{4} = \dfrac{5}{2}\) or \(x = \dfrac{-4}{4} = -1\).

(ii) \(5x^2 + 13x + 8 = 0\)

Identify: \(a = 5\), \(b = 13\), \(c = 8\).

Discriminant: \(D = 13^2 - 4(5)(8) = 169 - 160 = 9\).

Formula: \(x = \dfrac{-13 \pm \sqrt{9}}{2(5)} = \dfrac{-13 \pm 3}{10}\).

So, \(x = -1\) or \(x = -\dfrac{8}{5}\).

(iii) \(-3x^2 + 5x + 12 = 0\)

Multiply through by -1 to make \(a\) positive: \(3x^2 - 5x - 12 = 0\).

Now, \(a = 3\), \(b = -5\), \(c = -12\).

Discriminant: \(D = (-5)^2 - 4(3)(-12) = 25 + 144 = 169\).

Formula: \(x = \dfrac{5 \pm \sqrt{169}}{6} = \dfrac{5 \pm 13}{6}\).

So, \(x = 3\) or \(x = -\dfrac{4}{3}\).

(iv) \(-x^2 + 7x - 10 = 0\)

Multiply through by -1: \(x^2 - 7x + 10 = 0\).

Now, \(a = 1\), \(b = -7\), \(c = 10\).

Discriminant: \(D = (-7)^2 - 4(1)(10) = 49 - 40 = 9\).

Formula: \(x = \dfrac{7 \pm \sqrt{9}}{2} = \dfrac{7 \pm 3}{2}\).

So, \(x = 5\) or \(x = 2\).

(v) \(x^2 + 2\sqrt{2}\,x - 6 = 0\)

Here, \(a = 1\), \(b = 2\sqrt{2}\), \(c = -6\).

Discriminant: \(D = (2\sqrt{2})^2 - 4(1)(-6) = 8 + 24 = 32\).

\(\sqrt{32} = 4\sqrt{2}\).

Formula: \(x = \dfrac{-2\sqrt{2} \pm 4\sqrt{2}}{2}\).

So, \(x = \sqrt{2}\) or \(x = -3\sqrt{2}\).

(vi) \(x^2 - 3\sqrt{5}\,x + 10 = 0\)

Here, \(a = 1\), \(b = -3\sqrt{5}\), \(c = 10\).

Discriminant: \(D = (-3\sqrt{5})^2 - 4(1)(10) = 45 - 40 = 5\).

Formula: \(x = \dfrac{3\sqrt{5} \pm \sqrt{5}}{2}\).

Factor out \(\sqrt{5}\): \(x = \dfrac{\sqrt{5}(3 \pm 1)}{2}\).

So, \(x = 2\sqrt{5}\) or \(x = \sqrt{5}\).

(vii) \(\dfrac{1}{2}x^2 - \sqrt{11}\,x + 1 = 0\)

Multiply through by 2: \(x^2 - 2\sqrt{11}x + 2 = 0\).

Now, \(a = 1\), \(b = -2\sqrt{11}\), \(c = 2\).

Discriminant: \(D = (-2\sqrt{11})^2 - 4(1)(2) = 44 - 8 = 36\).

\(\sqrt{36} = 6\).

Formula: \(x = \dfrac{2\sqrt{11} \pm 6}{2} = \sqrt{11} \pm 3\).

So, \(x = \sqrt{11}+3\) or \(x = \sqrt{11}-3\).

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 4: Quadatric Equation – Exercise 4.3 | Detailed Answers