Quadratic equation in one unknown

Question

If the equation x²+2(a-1)x-(a²+4a-6ab+9b²)=0 has a double real root, where a and b are constants,find the values of a and b.

Answer 1

The discriminant is zero.

B^2 - 4AC = 0 = with A=1 , B = 2(a-1) and C = (a^2 + 4a - 6ab + 9b^2)

4 (a-1)^2 + (4)(1)(a^2 + 4a - 6ab + 9b^2) = 0 <--- the second sign is negative

4(a^2 - 2a + 1) + (4)(1)(a^2 + 4a - 6ab + 9b^2) = 0 <--- FOIL

a^2 - 2a + 1 + a^2 + 4a - 6ab + 9b^2 = 0 <--- divides by 4

a^2 + 2a + 1 + a^2 - 6ab + 9b^2 = 0 <--- combines the terms : -2a + 4a = 2a

(a+1)^2 + (a-3b)^2 = 0

Since the square of a number MUST be positive or zero,

the ONLY way this can happen is if a+1=0 and a-3b = 0

a= -1 and a = 3b

-1 = 3b

b= (-1/3)

THEN

B = 2(a-1) = 2(-1 -1) = 2(-2) = -4

C = a^2 + 4a - 6ab + 9b^2

= (-1)^2 + 4(-1) - 6(-1)(-1/3) + 9(-1/3)^2

= 1 - 4 - 2 + 1

= -4

The quadratic equation is

x^2 - 4x + 4 = 0

( x -2 )(x - 2 ) = 0

So x=2 is the double root