Question

Find the value(s) of k for which the equation

X^2+5kx+16 has real and equal roots.

Answer 1

for real and equal roots

D=0

D=b2-4ac

0=(5k)2-4(1)(16)

0=25k2-64

25k2=64

k2=64/25

$k = \sqrt{64 \div 25}$

k=8/5

k

Answer 2

HEY MATE!

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HERE IS YOUR ANSWER :

STEP BY STEP EXPLANATION:

In equation x^2+5Kx+16=0, a=1;b=5K;c=16

Discriminant =b^2-4ac

Discriminant =(5k)^2-4(1)(16)

Discriminant =25k^2-64

As the equation has real and equal roots so, discriminant is equal to 0

That is, 25K^2-64=0

25K^2=64

K^2=64/25

K=root=64/25

K=8/5

Thus , value of K is 8/5.

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HOPE IT HELPED ^_^