# If one root of the quadratic equation ax^2 + 8x + 12 = 0 is twice the other, what is the value of  a?

## Expert Answers We have the equation ax^2 + 8x + 12 = 0.

The roots of a quadratic equation ax^2 + bx + c = 0 are x1 = [-b + sqrt (b^2 – 4ac)]/2a and   x2 = [-b - sqrt (b^2 – 4ac)]/2a.

Here one of the roots is twice the other

=> [-b + sqrt (b^2 – 4ac)]/2a = 2*[-b - sqrt (b^2 – 4ac)]/2a

=> [-8 + sqrt (8^2 – 4*12*a)]/2a = 2*[-8 - sqrt (8^2 – 4*12*a)]/2a

=> [-8 + sqrt (8^2 – 4*12*a)] = [-16 – 2*sqrt (8^2 – 4*12*a)]

=> -8 + 16 + 3* sqrt (8^2 – 4*12*a) = 0

=> 8 + 3* sqrt (8^2 – 4*12*a) = 0

=> 3* sqrt (8^2 – 4*12*a) = -8

square both the sides

=> 9*(8^2 – 4*12*a) = 64

=> 9*64 – 9*48*a = 64

=> 9* 48* a = 64*8

=> a = 32/27

Therefore if a is 32/27 in the quadratic equation ax^2 + 8x + 12 = 0, one root is twice the other.

Approved by eNotes Editorial Team

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