Question

solve for k if equation
$2 {x}^{2} + kx + x + 8 = 0$
will have real and equal root​

Answer 1

Given Equation,

$\sf{2x {}^{2} + kx + x + 8 = 0} \\ \\ \implies \: \boxed{\sf{2x {}^{2} + (k + 1)x + 8 = 0 }}$

On comparing the boxed equation with ax²+bx+c = 0,

a = 2,b = k+1 and c = 8

As the equation has equal roots,D = 0

$\implies \: \sf{(k + 1) {}^{2} - 4(2)(8) = 0} \\ \\ \implies \: \sf{(k + 1) {}^{2} - 64 = 0 } \\ \\ \implies \: \sf{(k + 1) {}^{2} = 64} \\ \\ \implies \: \sf{k + 1 = \pm8} \\ \\ \implies \: \boxed{ \sf{k = 7 \: or \: - 9}}$

Answer 2

ANSWER:-

Given:

If equation 2x² +kx +x +8= 0

To find:

Solve for x.

Solution:

We can write above equation is form,

2x² + (k+1)x +8 =0

Now,

The roots of the quadratic equation

Ax² + Bx +c =0 by the compared.

⏺️A= 2

⏺️B= (k+1)

⏺️C= 8

So,

Discriminate value, D= 0

Therefore,

D= b² -4ac

=) (k+1)² -4× 2 ×8=0

=) (k+1)² - 64=0

=) (k+1)² = 64

=) (k+1)² = (8)²

=) k+1 = 8

=) k= 8 -1

=) k= 7

Hope it helps ☺️