Question

determine the positive value of k for which both tha equations x2+kx+64=0 and x2-8x+k=0 will have real roots

Answer 1

The positive value of k is 16.

Answer 2

Hi ,

x² + kx + 64 = 0 -----( 1 )

x² - 8x + k = 0 -------( 2 )

compare above equations with ax² + bx + c = 0,

1 ) a = 1 , b = k , c = 64

discreaminant = D ≥ 0

[ real roots ]

b² - 4ac ≥ 0

k² - 4 × 1 × 64 ≥ 0

k² ≥ 4 × 64

k ≥ ± 16

but k is a positive real value .

Therefore ,

k ≥ 16

2 ) a = 1 , b = -8 , c = k

D ≥ 0

( -8 )² - 4 × 1 × k ≥ 0

64 - 4k ≥ 0

-4k ≥ -64

k ≤ 64/4

k ≤ 16

Therefore ,

required positive which satisfy

both equations is 16

k = 16

I hope this helps you.

: )



= k² -