Is $x = - 2$ a solution of the equation $x^{2} - 2 x + 8 = 0$ ?

Open in App

Solution

Let,

$p (x) = x^{2} - 2 x + 8$

$⟹ p (- 2) = (- 2)^{2} - (- 2) 2 + 8$

$⟹ p (- 2) = 16$

Hence it is not the solution of given eq.

Suggest Corrections

Similar questions

{log}_{x^{2} + 6 x + 8} {log}_{{2 x}^{2} + 2 x + 3} (x^{2} - 2 x) = 0

$x = - 4$ is a solution to the equation $2 x + 8 = 0$

x^{2} - (tan \frac{a}{2} + cot \frac{a}{2}) x + tan \frac{a}{2} \times cot \frac{a}{2} = 0

x^{2} - 2 x - 8 = 0

x

(x - 2 y) \frac{d y}{d x} + 2 x + y = 0

Join BYJU'S Learning Program