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For what value of k, $(4 - k) x^{2} + (2 k + 4) x + (8 k + 1) = 0$ is a perfect quadratic equation.

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0, 1

1, 1

2, 2

0, 3

answer is D.

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Detailed Solution

Given,

(4 - k) x^{2} + (2 k + 4) x + (8 k + 1) = 0

For perfect square the value of discriminant is equal to 0. So,

D = 0

b^{2} - 4 ac = 0

Now we have,

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

Put the values,

(2 k + 4)^{2} - 4 (4 - k) (8 k + 1) = 0

4 k^{2} + 16 k + 16 - 4 (32 k + 4 - 8 k^{2} - k) = 0

4 k^{2} + 16 k + 16 - 4 (31 k + 4 - 8 k^{2}) = 0

4 k^{2} + 16 k + 16 - 124 k - 16 + 32 k^{2} = 0

36 k^{2} + 16 k - 124 k = 0

36 k^{2} - 108 k = 0

36 k (k - 3) = 0

k = 0, 3

So, values of k are 0 and 3.
Correct option is 4.

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