Question

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

Answer 1

Step-by-step explanation:

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

(4x2-2xk)+(2x2+4x)+8x+1=0

4x2-2xk+2x2+4x+8x+1=0

4x2-2xk+2x2+12x+1=0

2xk=-4x2-2x2+12x+1

2xk=-6x2+12x+1

k=-6x2+12x+1/2x

k=-6x2+6x+1

So, the value of k is (- 6x2+6x+1).

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Answer 2

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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.