Q. 31

# Solve the following quadratic equations by factorization: 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(y2 + 1) = y(c2+1)

cy2 + c = c2y + y

cy2 – c2y –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 ⇒ x2 - bx - ax + ab = a/b

⇒ b(x2 - bx - ax + ab) = a

⇒ bx2 - b2x - abx + ab= 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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