Question

the value of k for which the equation 2x^2-kx+x+8=0 will have real and equal roots​

Answer 1

Answer:

9 and -7

Step-by-step explanation:

2x^2+x(-k+1)+8=0

the equation will have real and equal roots if b^2 -4ac=0

Here, b=(-k+1),a=2,c=8

b^2-4ac=0

(1-k)^2-4*2*8=0

(1+k^2-2k)-64=0

k^2-2k-63=0

(k-9)(k+7)=0

So,

k=9 or k=-7

Answer 2

Answer:

Step-by-step explanation:

2x ^2 - k x + x +8 =0

2 x ^2  -k x + x = -8

x * x -kx +x = -8 / 2

                    = -4

2x * x - kx = -4

x* x -k x = -4 / 2

             = -2