Question

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

5x² - 4x + 2 + k(4x² - 2x - 1) = 0

Answer 1

The given quadric equation is 5x² - 4x + 2 + k(4x² - 2x - 1) = 0, and roots are real and equal

Then find the value of k.

Here,

5x² - 4x + 2 + k(4x² - 2x - 1) = 0

5x² - 4x + 2 + 4kx² - 2kx - k = 0

(5 + 4k)x² - (4 + 2k)x + (2 - k) = 0

So,

a = (5 + 4k), b = -(4 + 2k) and c = (2 - k)

As we know that D = b² - 4ac

Putting the value of 

a = (5 + 4k), b = -(4 + 2k) and c = (2 - k)

= {-(4 + 2k)}² - 4 x (5 + 4k) x (2 - k)

= (16 + 16k + 4k²) - 4(10 + 3k - 4k²)

= 16 + 16k + 4k² - 40 - 12k + 16k²

= 20k² + 4k - 24

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

Thus,

20k² + 4k - 24 = 0

4(5k² + k - 6) = 0

5k² + k - 6 = 0

Now factorizing of the above equation

5k² + k - 6 = 0

5k² + 6k - 5k - 6 = 0

k(5k + 6) - 1(5k + 6) = 0

(5k + 6)(k - 1) = 0

So, either

5k + 6 = 0

5k = -6

k = -6/5

Or

k - 1 = 0

k = 1

Therefore, the value of k = 1, -6/5