Question

In each of the examples given below determine whether the values given against each of the quadratic equation are the roots of the equation or not.

(i) x² + 3x $-$ 4 = 0, x = 1, – 2, –3

Answer 1

Given: x² + 3x – 4 = 0 and x = 1, – 2, –3

On substituting x = 1 in L.H.S. of the given equation, we get:
(1)² + 3(1) – 4
= 1 + 3 – 4
= 0
L.H.S. = R.H.S.
So, x = 1 satisfies the given equation.
Therefore, x = 1 is a root of the given quadratic equation.

On substituting x = –2 in L.H.S. of the given equation, we get:
(–2)² + 3(–2) – 4
= 4 – 6 – 4
= –6
$\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.$
Thus, x = –2 does not satisfy the given equation.
Therefore, x = –2 is not a root of the given quadratic equation.

On substituting x = –3 in L.H.S. of the given equation, we get:
(–3)² + 3(–3) – 4
= 9 – 9 – 4
= – 4
$\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.$
Thus, x = –3 does not satisfy the given equation.
Therefore, x = –3 is not a root of the given quadratic equation.