NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 4: Quadatric Equation

Exercise 4.3

We will solve all problems using the quadratic formula:

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

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 \(x = \sqrt{11}-3\).

(i)

Equation: \(2x^2 + \dfrac{5}{3}x - 2 = 0\)

Step 1: Remove the fraction by multiplying everything by 3:

\(6x^2 + 5x - 6 = 0\)

Step 2: Multiply first and last coefficient: \(6 \times -6 = -36\).

We need two numbers whose product is -36 and sum is +5. These are +9 and -4.

Step 3: Split middle term:

\(6x^2 + 9x - 4x - 6 = 0\)

Step 4: Group terms:

\((6x^2 + 9x) - (4x + 6) = 0\)

Step 5: Factor each group:

\(3x(2x + 3) - 2(2x + 3) = 0\)

Step 6: Take common factor:

\((2x + 3)(3x - 2) = 0\)

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

(ii)

Equation: \(\dfrac{2}{5}x^2 - x - \dfrac{3}{5} = 0\)

Step 1: Remove fraction by multiplying by 5:

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

Step 2: Multiply first and last coefficient: \(2 \times -3 = -6\).

We need two numbers with product -6 and sum -5. These are -6 and +1.

Step 3: Split middle term:

\(2x^2 - 6x + x - 3 = 0\)

Step 4: Group terms:

\((2x^2 - 6x) + (x - 3) = 0\)

Step 5: Factor:

\(2x(x - 3) + 1(x - 3) = 0\)

Step 6: Take common factor:

\((2x + 1)(x - 3) = 0\)

So, \(x = -\tfrac{1}{2}\) or \(x = 3\).

(iii)

Equation: \(3\sqrt{2}x^2 - 5x - \sqrt{2} = 0\)

Step 1: Multiply first and last coefficient: \(3\sqrt{2} \times -\sqrt{2} = -6\).

We need two numbers with product -6 and sum -5. These are -6 and +1.

Step 2: Split middle term:

\(3\sqrt{2}x^2 - 6x + x - \sqrt{2} = 0\)

Step 3: Group terms:

\((3\sqrt{2}x^2 - 6x) + (x - \sqrt{2}) = 0\)

Step 4: Factor:

\(3x(\sqrt{2}x - 2) + 1(x - \sqrt{2}) = 0\)

Step 5: Rearrange grouping:

\((3\sqrt{2}x + 1)(x - \sqrt{2}) = 0\)

So, \(x = \sqrt{2}\) or \(x = -\tfrac{\sqrt{2}}{6}\).

(iv)

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

Step 1: Look for possible factors with surds (square roots).

Try: \((3x - \sqrt{5})(x + 2\sqrt{5})\).

Step 2: Expand to check:

\(3x^2 + 6x\sqrt{5} - x\sqrt{5} - 10\)

= \(3x^2 + 5\sqrt{5}x - 10\). Correct!

Step 3: So equation is \((3x - \sqrt{5})(x + 2\sqrt{5}) = 0\).

Hence, \(x = \tfrac{\sqrt{5}}{3}\) or \(x = -2\sqrt{5}\).

(v)

Equation: \(21x^2 - 2x + \dfrac{1}{21} = 0\)

Step 1: Multiply everything by 21 to remove fraction:

\(441x^2 - 42x + 1 = 0\)

Step 2: Notice this looks like a perfect square.

\((21x - 1)^2 = 441x^2 - 42x + 1\)

Step 3: So equation is \((21x - 1)^2 = 0\).

That means root is \(x = \tfrac{1}{21}\), repeated twice (double root).