Question

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

Answer 1

hello dear,

b2-4ac=0(given)

(2k+4)2-4(4-k)(8k+1)=0

4k2+16k+16-4(32k+4-8k2-k)=0

4k2+16k+16-128k-16+32k2+4k=0

36k2-108k=0 (Dividing the equation by 4)

9k2-27k=0

9k2=27k

9k=27k/k

9k=27

K=27/9

K=3

hope it will will help you. :-)

Answer 2

$\large\bf{\underline\green{☕︎Good \: Morning☕︎}}$

Given, (4 – k)x² + (2k + 4)x + (8k + 1) = 0

It is in the form of ax² + bx + c = 0

Where, a = 4 – k, b = 2k + 4, c = 8k + 1

Calculating the discriminant,

D = b²– 4ac

= (2k + 4)² – 4(4 – k)(8k + 1)

= 4k² + 16 + 4k – 4(32 + 4 – 8k² – k)

= 4(k² + 4 + k – 32 – 4 + 8k² + k)

= 4(9k² – 27k)

As the given equation is a perfect square, then D = 0

⇒ 4(9k² – 27k) = 0

⇒ (9k² – 27k) = 0

⇒ 3k(k – 3) = 0

Thus, 3k = 0

⇒ k = 0 Or k-3 = 0

⇒k = 3

Hence, the value of k should be 0 or 3 for the given to be perfect square.