Solve the following quadratic equation for x:
x2 – 4ax – b2 + 4a2 = 0
x2 – 4ax – b2 + 4a2 = 0
x2 – 4ax – (b2 - 4a2) = 0
x2 – 4ax – [(b)2 – (2a)2] = 0
x2 – 4ax – (b – 2a)(b + 2a) = 0
x2 + (b – 2a – b – 2a) x – (b – 2a)(b + 2a) = 0
x2 + (b – 2a) x + ( - b – 2a) x - (b – 2a)(b + 2a) = 0
x2 + (b – 2a) x - (b + 2a) x - (b – 2a)(b + 2a) = 0
x [x + (b - 2a)] - (b + 2a)[x + (b - 2a)] = 0
[x + (b – 2a)][x – (b + 2a)] = 0
x + (b – 2a) = 0 or x – (b + 2a) = 0
x = - (b – 2a) or x = (b + 2a)
Thus the solutions for the given equation are;
x = - (b – 2a) and
x = (b + 2a)
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