# What is the value of b in the quadratic equation 4x^2 + bx + 12 =0 so that one root is thrice the other.

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### 2 Answers

To find the value of b in 4x^2+bx+12 = 0, given that one root is 3 times the other.

Let a and 3a be the roots.

Therefore by the relation between the roots and coeficients, we have

a+3a = -b/4 .Or

4a = -b/12...............(1)

a*3a = 12/4. Or

a^2 = 1.................(2).

Eliminate a between (1) and (2) :

(-b/12*4)^2 = 1.

b^2/48^2 = 1

b^2 = 48^2

We take square root:

b = 48

b = -48

Therefore the values of b are 96 or -96.

Let us take one of the roots as R, the other root is 3R.

Also, let’s express 4x^2 + bx + 12 =0 as an equivalent form x^2 +(b/4)x +3 =0

Now (x-R)(x-3R) = x^2 + (b/4)x +3

=> x^2 – 4Rx + 3R^2 = x^2 + (b/4)x + 3

So 3R^2 = 3 => R^2 = 1 => R = 1, -1

Also b = -16R

**So b can be -16 or 16**

To verify:

for 4x^2 – 16 + 12 =0

=> x^2 – 4x +3 =0

=> x^2 – 3x – x +3 =0

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

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

The roots are 1 and 3.

for 4x^2 + 16 + 12 =0

=> x^2 + 4x +3 =0

=> x^2 + 3x + x +3 =0

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

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

The roots are -1 and -3.