#### Question

If the sum of the roots of the equation \[x^2 - \left( k + 6 \right)x + 2\left( 2k - 1 \right) = 0\] is equal to half of their product, then k =

6

7

1

5

#### Solution

The given quadric equation is `x^2 - (k+6)x + 2 (2k - 1) = 0 `, and roots are equal

Then find the value of *k.*

Let `alpha and beta ` be two roots of given equation

And, a = 1, b = -(k + 6) and , c = 2 (2k - 1)

Then, as we know that sum of the roots

`alpha + beta = (-b)/a`

`alpha + beta = (-{-(k + 6)})/1`

`= (k + 6)`

And the product of the roots

`alpha . beta = c /a`

`alphabeta = (2(2k - 1))/1`

` = 2 (2k- 1)`

According to question, sum of the roots ` = 1/2 xx` product of the roots

`(k + 6) = 1/2 xx 2 (2k - 1)`

`k+6 = 2k - 1`

` 6+1 = 2k - k`

7 = k

Therefore, the value of **k = 7.**