Q. 434.1( 10 Votes )

Solve each of the following quadratic equations:

x2 - 2ax - (4b2 - a2) = 0

Answer :

x2 - 2ax - (4b2 - a2) = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:


Product = a.c


For the given equation a = 1 b = - 2a c = - (4b2 - a2)


= 1. - (4b2 - a2)


= - (4b2 - a2)


And either of their sum or difference = b


= - 2a


Thus the two terms are (2b - a) and - (2b + a)


Difference = 2b - a - 2b - a


= - 2a


Product = (2b - a) - (2b + a)


( using a2 - b2 = (a + b) (a - b))


= - (4b2 - a2)


x2 - 2ax - (4b2 - a2) = 0


x2 + (2b - a)x - (2b + a)x - (2b - a)(2b + a) = 0


x[x + (2b - a)] - (2b + a)[x + (2b - a)] = 0


[x + (2b - a)] [x - (2b + a)] = 0


[x + (2b - a)] = 0 or [x - (2b + a)] = 0


x = (a - 2b) or x = (a + 2b)


Hence the roots of given equation are (a - 2b) or x = (a + 2b)


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