Q. 31

# Solve the following quadratic equations by factorization:

Answer :

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

Let (x – a)/(x – b) = y and a/b = c ..... (1)

So (x – b)/(x – a) = 1/ y and b/a = 1/c .... (2)

⇒ c(y^{2} + 1) = y(c^{2}+1)

⇒ cy^{2} + c = c^{2}y + y

⇒ cy^{2} – c^{2}y –y + c = 0

⇒ cy(y – c) – 1(y – c) = 0

⇒ (cy – 1)(y – c) = 0

⇒ y = 1/c, c

From (1)

⇒ (x – a)/(x – b) = c and a/b = c

⇒ x^{2} - bx - ax + ab = a/b

⇒ b(x^{2} - bx - ax + ab) = a

⇒ bx^{2} - b^{2}x - abx + ab^{2 }= a

⇒ bx ( x - b ) - ab ( x - b ) = a

⇒ (bx - ab) ( x - b ) = a

⇒ (bx - ab) = a and ( x - b ) = a

So,

bx - ab = a

⇒ bx = a + ab

⇒ x =( a + ab ) / b

And

x - b = a

⇒ x = a + b

Hence x =( a + ab ) / b , (a+b)

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KC Sinha - Mathematics

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