Q. 15 A5.0( 2 Votes )

Solve the followi

Answer :

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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