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

Answer:

The value of k is $\frac{-3}{5}$

The roots are -1 and $\frac{1}{3}$

Step-by-step explanation:

In any quadratic equation $ax^{2}+bx+c=0$ the sum of its roots is $\frac{-b}{a}$ and the product of its root is $\frac{c}{a}$

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

The sum is $\frac{--(1+k)}{k}=\frac{1+k}{k}$

The product is $\frac{3k+2}{k}$

∵ The sum of the roots is twice their product

∴ $\frac{k+1}{k}=2[\frac{3k+2}{k}]$

∴ $\frac{1+k}{k}=\frac{6k+4}{k}$⇒Multiply both sides by k

1 + k = 6k + 4⇒ 1 - 4 = 6k - k

5k = -3 ⇒ $k=\frac{-3}{5}$

Use the value of k in the equation:

$\frac{-3}{5}x^{2}-[1+\frac{-3}{5}]x+[(3)(\frac{-3}{5})+2]=0$

$\frac{-3}{5}x^{2}-\frac{2}{5}x+\frac{1}{5}=0$⇒ 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