Question

For what value of k, (4-k)x2 + (2k + 4)x + (8k + 1) = 0, is a perfect square.

Answer 1

The given quadratic will be a perfect square if it has two real and equal roots. To have two real and equal roots the discriminant must be zero.
i.e., b²-4ac=0
or, (2k+4)²-4(4-k)(8k+1)=0
or, 4k²+16k+16-4(32k-8k²+4-k)=0
o², 4k²+16k+16-128k+32k²-16+4k=0
or, 36k²-108k=0
or, 36k(k-3)=0
either 36k=0
or, k=0
or, k-3=0
or, k=3
∴, k=0,3 Ans.

Answer 2

Answer:

The value of k is 3 and 0

Step-by-step explanation:

Given :-

• for perfect square
• b2 - 4ac =0

To find :-

• the value of k =?

Solution :-

• (2k+4)2 -4 (4 - k)(8k+1) =0
• 4k2 +16k+16 - 4 (32k -8k2 + 4 -k) =0
• 4k2 +16k +16 -128k +32k2 -16 +4k =0
• 36k2 -108k =0
• 36k (k -3) =0
• 36k =0
• k =0
• or, k -3 =0
• k =3

therefore the value of k is 3 and 0