How do you solve #x^2=24x+10# by completing the square?

2 Answers
Sep 25, 2016

Answer:

#color(green)(x=12-sqrt(154))color(white)("XX")# or #color(white)("XX")color(green)(x=12+sqrt(154)#

Explanation:

Given
#color(white)("XXX")x^2=24x+10#

Shift all the terms which include the variable #x# to the left side:
#color(white)("XXX")x^2-24x=10#

Since, in general, the expansion of a squared binomial has the structure:
#color(white)("XXX")(x+a)^2=underline(x^2+2ax) + a^2#

If #x^2-24x# are the first two terms of the expansion of a squared binomial
then the third terms should be #(-24/2)^2=(-12)^2=12^2#

In order to complete the square we will need to add #12^2# (to both sides)

#color(white)("XXX")x^2-24x+12^2=10+12^2#

Writing as a squared binomial and simplifying:
#color(white)("XXX")(x-12)^2=154#

Taking the square root of both sides
#color(white)("XXX")x-12=+-sqrt(154)#
Adding #12# to both sides
#color(white)("XXX")x=12+-sqrt(154)#

Answer:

#x=12+-sqrt(154)#

Explanation:

#x^2-24x = 10#
take the factor of x, half it and square, then add both sides
#x^2-24x+144 = 10+144#
#(x-12)^2 = 154#
#sqrt((x-12)^2)=+-sqrt(154)#
#(x-12)=+-sqrt(154)#
#x=12+-sqrt(154)#