Question

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

Answer 1

Answer:

SOLUTION :  

Given :  (4 - k) x² + (2k+ 4)x + (8k + 1) = 0

On comparing the given equation with ax² + bx + c = 0  

Here, a = 4 - k , b = 2k + 4 , c = 8k + 1

D(discriminant) = b² – 4ac

D = (2k + 4)² - 4(4 - k)(8k + 1)

D = [(2k)² + 4² + 2 × 2k × 4 ) - 4(32k + 4 - 8k² - k)

[(a + b)² = a² + b² + 2ab]

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

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

= 4[k²  + 4 + 4 k - 32k + k - 4 + 8k² ]

= 4 [k² + 8k² + 4k + k - 32k  + 4 - 4]

= 4 [9k² -  27k]

Since, the given equation is a perfect square

Therefore D = 0

4 [9k² -  27k] = 0

[9k² -  27k] = 0

3k (k - 3) = 0

3k = 0  or (k - 3) = 0

k = 0 or  k  = 3

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

Step-by-step explanation: