Question

for the equation kx^2-(1+k)x+(3k+2)=0, the sum of the roots is twice their product. find k and the 2 roots​

Answer 1

The value of k is
`(-3)/(5)`

The roots are -1 and
`(1)/(3)`

Explanation:

In any quadratic equation
`ax^(2)+bx+c=0` the sum of its roots is
`(-b)/(a)` and the product of its root is
`(c)/(a)`

In the equation
`kx^(2)-(1+k)x+(3k+2)=0`

The sum is
`(--(1+k))/(k)=(1+k)/(k)`

The product is
`(3k+2)/(k)`

∵ The sum of the roots is twice their product

∴
`(k+1)/(k)=2[(3k+2)/(k)]`

∴
`(1+k)/(k)=(6k+4)/(k)`⇒Multiply both sides by k

1 + k = 6k + 4⇒ 1 - 4 = 6k - k

5k = -3 ⇒
`k=(-3)/(5)`

Use the value of k in the equation:

`(-3)/(5)x^(2)-[1+(-3)/(5)]x+[(3)((-3)/(5))+2]=0`

`(-3)/(5)x^(2)-(2)/(5)x+(1)/(5)=0`⇒ Multiply equation by 5

`-3x^(2)-2x+1=0`⇒ Multiply equation by -1

`3x^(2)+2x-1=0`⇒ use factorization to find roots

(3x - 1)(x + 1) = 0

3x -1 = 0⇒ 3x = 1⇒ x = 1/3

x + 1 = 0⇒ x = -1

The roots are 1/3 and -1