How do you solve # 21x^2 + 22x - 24 = 0#?

1 Answer
Feb 26, 2017

Answer:

#x=-12/7# or #x=2/3#

Explanation:

If the discriminant of a quadratic equation #ax^2+bx+c=0#, which is #Delta=b^2-4ac# is a complete square of a rational number, we can solve it by factorization.

Here in the given equation #21x^2+22x-24=0#, #Delta=22^2-4xx21xx(-24)=484+2016=2500=50^2#, hence we can solve it by factorization.

For factorization, split middle term #b=22# in two parts whose product is #ac=-21xx24=-504#. In other words two numbers, whose difference is #22# and product is #504#. These are are #-14# and #36#.

Therefore, #21x^2+22x-24=0# can be written as

#21x^2-14x+36x-24=0#

or #7x(3x-2)+12(3x-2)=0#

or #(7x+12)(3x-2)=0#

Hence either #7x+12=0# i.e. #x=-12/7#

or #3x-2=0# i.e. #x=2/3#