Question

For what value of k, for which the roots of the equation:
$(4 + k) {x}^{2} + (2k + 4)x + (8k + 1) = 0$
are real and equal?​

Answer 1

Answer:

For what value of k, for which the roots of the equation:

$(4 + k) {x}^{2} + (2k + 4)x + (8k + 1) = 0$

are real and equal?

Answer 2

Answer:

k=0 or -29/7

Step-by-step explanation:

(4+k)²+(2k+4)k+8k+1=0

here a=4+k,b=2k+4,c=8k+1

if roots are equal then

b²=4ac

(2k+4)²=4(4+k)(8k+1)

4k²+16k+16=4( 32k+4+8k²+k)

4k²+16k+16=128k+16+32k²+4k

28k²+116k=0

4k(7k+29)=0

k=0

or

7k+29=0

k=-29/7