#### Question

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

#### Solution

The given quadric equation is (4 - k)x^{2} + (2k + 4)x + (8k + 1) = 0, and roots are real and equal

Then find the value of *k.*

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

As we know that D = b^{2} - 4ac

Putting the value of a = (4 - k), b = (2k + 4) and c = (8k + 1)

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

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

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

= 36k^{2} - 108k + 0

= 36k^{2} - 108k

The given equation will have real and equal roots, if D = 0

Thus,

36k^{2} - 108k = 0

18k(2k - 6) = 0

k(2k - 6) = 0

Now factorizing of the above equation

k(2k - 6) = 0

So, either

k = 0

or

2k - 6 = 0

2k = 6

k = 6/2 = 3

Therefore, the value of k = 0, 3.

Is there an error in this question or solution?

#### APPEARS IN

Solution For What Value of K, (4 - K)X2 + (2k + 4)X + (8k + 1) = 0, is a Perfect Square. Concept: Nature of Roots.