Question

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

(2k + 1)x² + 2(k + 3)x + (k + 5) = 0

Answer 1

The given quadric equation is (2k + 1)x² + 2(k + 3)x + k + 5 = 0, and roots are real and equal

Then find the value of k.

Here,

a = (2k + 1), b = 2(k + 3) and c = k + 5

As we know that D = b² - 4ac

Putting the value of a = (2k + 1), b = 2(k + 3) and c = k + 5

={2(k + 3)}² - 4 x (2k + 1) x (k + 5)

= {4(k² + 6k + 9)} - 4(2k² + 11k + 5)

= 4k² + 24k + 36 - 8k² - 44k - 20

= -4k² - 20k + 16

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

-4k² - 20k + 16 = 0

-4(k² + 5k - 4) = 0

k² + 5k - 4 = 0

Now factorizing the above equation

k² + 5k - 4

`k=(-b+-sqrt(b^2-4ac))/(2a)`

`k=(-5+-sqrt(25+16))/2`

`k=-5+-sqrt41/2`

So, either

Therefore, the value of `k=-5+-sqrt41/2`