How do you solve #4x^2 - 12x + 9 = 0#?

1 Answer
Jun 23, 2016

Answer:

#x=3/2#

Explanation:

In #4x^2-12x+9=0# as niddle term #-12x# is twice the product (with negative sign) of square roots of first (#sqt(4x^2)=2x#) and third term (#sqrt9=3#,

it can be factorized using formula #a^2-2ab+b^2=(a-b)^2#

Hence #4x^2-12x+9=0#

#hArr(2x)^2-2xx3xx3x+3^2=0#

or #(2x-3)^2=0#

and hence #2x=3# or #x=3/2#