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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​

1 Answer

1 vote

Answer:

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

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