Answer:
The value of k is [tex]\frac{-3}{5}[/tex]
The roots are -1 and [tex]\frac{1}{3}[/tex]
Step-by-step explanation:
In any quadratic equation [tex]ax^{2}+bx+c=0[/tex] the sum of its roots is [tex]\frac{-b}{a}[/tex] and the product of its root is [tex]\frac{c}{a}[/tex]
In the equation [tex]kx^{2}-(1+k)x+(3k+2)=0[/tex]
The sum is [tex]\frac{--(1+k)}{k}=\frac{1+k}{k}[/tex]
The product is [tex]\frac{3k+2}{k}[/tex]
∵ The sum of the roots is twice their product
∴ [tex]\frac{k+1}{k}=2[\frac{3k+2}{k}][/tex]
∴ [tex]\frac{1+k}{k}=\frac{6k+4}{k}[/tex]⇒Multiply both sides by k
1 + k = 6k + 4⇒ 1 - 4 = 6k - k
5k = -3 ⇒ [tex]k=\frac{-3}{5}[/tex]
Use the value of k in the equation:
[tex]\frac{-3}{5}x^{2}-[1+\frac{-3}{5}]x+[(3)(\frac{-3}{5})+2]=0[/tex]
[tex]\frac{-3}{5}x^{2}-\frac{2}{5}x+\frac{1}{5}=0[/tex]⇒ Multiply equation by 5
[tex]-3x^{2}-2x+1=0[/tex]⇒ Multiply equation by -1
[tex]3x^{2}+2x-1=0[/tex]⇒ 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
