How do you solve #2x ^ { 2} - 13x + 18- ( x - 2) ^ { 2}#?

1 Answer
Aug 7, 2017

Answer:

#(x-2)(x-7)#

If we presume that the equation equals zero, then:

#x=2# or #x=7#

Explanation:

#2x^2-13x+18-(x-2)^2#

We first expand the term #(x-2)^2#.

#2x^2-13x+18-[x(x-2)-2(x-2)]#

#2x^2-13x+18-[x^2-2x-2x+4]#

#2x^2-13x+18-[x^2-4x+4]#

We now open the brackets, changing the signs appropriately.

#2x^2-13x+18-x^2+4x-4#

Group all like terms with their preceding signs and then simplify.

#2x^2-x^2-13x+4x+18-4#

#x^2-9x+14#

Factorise.

#x^2-7x-2x+14#

#x(x-7)-2(x-7)#

#(x-2)(x-7)#

If we presume that the equation equals zero, then:

#x-2=0# and #x-7=0#

#x=2# or #x=7#