How do you solve #bx^2-b = x-b^2x# by factoring?

Question

1476 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-25T13:20:43

#color(green)( x = -b or x = 1/b#

We can solve for #x# by making Groups

#bx^2-b = x-b^2x#

Transposing the terms on the right to the left, we get:

# => bx^2 + b^2x - x - b = 0 #

Next, we form groups of 2 terms:

#=> (bx^2 + b^2x) - (x + b) = 0 #

Then we take the common factors out from each group

#=> bx(x+b) - 1(x+b) = 0#

The factor #x + b# is common to both the terms:

#=> (x + b)(bx - 1) = 0#

This tells us that :

#x + b = 0 or bx - 1 = 0#

#color(green)( x = -b or x = 1/b#