If one root of the quadratic equation $2 x^{2} + k x - 6 = 0$ is 2, find the value of k. Also, find the other root.
[2 MARKS]

Open in App

Solution

Concept : 1 Mark
Application :1 Mark

Since $x = 2$ is a root of the equation

$2 x^{2} + k x - 6 = 0$

$∴ 2 \times 2^{2} + 2 k - 6 = 0$

$\Rightarrow 8 + 2 k - 6 = 0 \Rightarrow 2 k + 2 = 0 \Rightarrow k = - 1$

Substituting $k = - 1$ in the equation $2 x^{2} + k x - 6 = 0$ , we get

$2 x^{2} - x - 6 = 0$

$\Rightarrow 2 x^{2} - 4 x + 3 x - 6 = 0$

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

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

$\Rightarrow x - 2 = 0, 2 x + 3 = 0 \Rightarrow x = 2, x = - \frac{3}{2}$

Hence, the other root is $- \frac{3}{2}$

Suggest Corrections

Similar questions

2 x^{2} + k x - 2 = 0

- 2

k

If one root of the quadratic equation $2 x^{2} + k x - 6 = 0$ is 2, then the value of k is

\frac{- 1}{2}

2 x^{2} + k x + 1 = 0

x

(K + 2) x^{2} - K x + 6 = 0.

k

2

2 x^{2} - k x + k = 0.

Join BYJU'S Learning Program