Q. 443.9( 8 Votes )

# Solve each of the following quadratic equations:

x^{2} - (2b - 1)x + (b^{2} - b - 20) = 0

Answer :

x^{2} - (2b - 1)x + (b^{2} - 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 = b^{2} - b - 20

= 1(b^{2} - b - 20)

= (b^{2} - 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)

= b^{2} - b - 20

x^{2} - (2b - 1)x + (b^{2} - b - 20) = 0

⇒ x^{2} - (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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