Q. 443.9( 8 Votes )

Solve each of the following quadratic equations:

x2 - (2b - 1)x + (b2 - b - 20) = 0

Answer :

x2 - (2b - 1)x + (b2 - b - 20) = 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 = - (2b - 1); c = b2 - b - 20


= 1(b2 - b - 20)


= (b2 - b - 20)


And either of their sum or difference = b


= - (2b - 1)


Thus the two terms are - (b - 5) and - (b + 4)


Sum = - (b - 5) - (b + 4)


= - b + 5 - b - 4


= - 2b + 1


= - (2b - 1)


Product = - (b - 5) - (b + 4)


= (b - 5) (b + 4)


= b2 - b - 20


x2 - (2b - 1)x + (b2 - b - 20) = 0


x2 - (b - 5)x - (b + 4)x + (b - 5)(b + 4) = 0


x[x - (b - 5)] - (b + 4)[x - (b - 5)] = 0


[x - (b - 5)] [x - (b + 4)] = 0


[x - (b - 5)] = 0 or [x - (b + 4)] = 0


x = (b - 5) or x = (b + 4)


Hence the roots of equation are (b - 5) or (b + 4)


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