Question

x² + 5x – 14 = 0, x = $\sqrt{2}$ , – 7, 3

Answer 1

Given: x² + 5x – 14 = 0 and x = $\sqrt{2}$ , – 7, 3

On substituting x = $\sqrt{2}$ in L.H.S. of the given equation, we get:
($\sqrt{2}$ )² + 5($\sqrt{2}$ ) – 14
= 2 + 5$\sqrt{2}$ – 14
= –12 + 5$\sqrt{2}$
$\Rightarrow$ L.H.S. $\ne$ R.H.S.
Thus, x = $\sqrt{2}$ does not satisfy the given equation.
Therefore, x = $\sqrt{2}$ is not a root of the given quadratic equation.

On substituting x = –7 in L.H.S. of the given equation, we get:
(–7)² + 5(–7) – 14
= 49 – 35 – 14
= 49 – 49
= 0
$\Rightarrow$ L.H.S. = R.H.S.
Thus, x = –7 satisfies the given equation.
Therefore, x = –7 is a root of the given quadratic equation.

On substituting x = 3 in L.H.S. of the given equation, we get:
(3)² + 5(3) – 14
= 9 + 15 – 14
= 24 – 14
= 10
$\Rightarrow$ L.H.S. $\ne$ R.H.S.
Thus, x = 3 does not satisfy the given equation.
Therefore, x = 3 is not a root of the given quadratic equation.