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)

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Quiz | Knowing the Nature of Roots44 mins
Take a Dip Into Quadratic graphs32 mins
Foundation | Practice Important Questions for Foundation54 mins
Nature of Roots of Quadratic Equations51 mins
Getting Familiar with Nature of Roots of Quadratic Equations51 mins
Understand The Concept of Quadratic Equation45 mins
Quiz | Lets Solve Imp. Qs of Quadratic Equation43 mins
Balance the Chemical Equations49 mins
Champ Quiz | Quadratic Equation33 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses