Question

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

Answer 1

According to quition

The above quadetic equation has equal roots

So b^2–4ac=0

(2k+4)^2–4(4-k)(8k+1)=0

4k^2+16+16k-(16–4k)(8k+1)=0

4k^2+16+16k=128k+16–32k^2–4k

36k^2–108=0

k(36k-108)=0

k=0 or k=3

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.