Question

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

Answer 1

$\huge{\mathtt{Answer}}$

We have,

$\sf{x {}^{2} + kx + 64 = 0 }$

Given that both the equations have equal roots. i.e, Discriminant is equal to zero

On comparing equation (1) with ax²+bx+c=0,

a=1,b=k and c=64

★As D=0,

→b² - 4ac=0

→b²=4ac

→k²=4(64)(1)

→k²=256

→k=16

Consider equation (2),

$\sf{x {}^{2} - 8x + k = 0 }$

On comparing with ax²+bx+c=0,

a=1,b= -8 and c=k

Now,

★D=0

→b²-4ac=0

→(-8)²=4(1)(k)

→4k=64

→k=16

Thus,in both the cases k=16