Question

Find the values of k for which the roots are real and equal in each of the following equation:

kx² + kx + 1 = -4x² - x

Answer 1

The given quadric equation is kx² + kx + 1 = -4x² - x, and roots are real and equal

Then find the value of k.

Here,

kx² + kx + 1 = -4x² - x

4x² + kx² + kx + x + 1 = 0

(4 + k)x² + (k + 1)x + 1 = 0

So,

a = (4 + k), b = (k + 1) and c = 1

As we know that D = b² - 4ac

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

= (k + 1)² - 4 x (4 + k) x (1)

= (k² + 2k + 1) - 16 - 4k

= k² - 2k - 15

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

Thus,

k² - 2k - 15 = 0

Now factorizing of the above equation

k² - 2k - 15 = 0

k² - 5k + 3k - 15 = 0

k(k - 5) + 3(k - 5) = 0

(k - 5)(k + 3) = 0

So, either

k - 5 = 0

k = 5

Or

k + 3 = 0

k = -3

Therefore, the value of k = 5, -3.