Question

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

Answer 1

Answer:

answer is 3

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Step-by-step explanation:

(4-K)X²+(2K+4)X+(8K+1)=0 is a perfect square when discriminant is zero.

d=b² - 4ac

a = 4-k

b = 2k + 4

c = 8k + 1

Value of K for which d is zero will be:

(2k+4)² - 4(4-k)(8k+1) = 0

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

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

36k² -108k+ 0 = 0

36k² = 108k

36k = 108

180/36 =3 =k