How do you factor # x ^ { 2} - 10x = - 11#?

1 Answer
Andy Y.
Nov 9, 2017

#x^2-10x+11=(x-5-sqrt(14))(x-5+sqrt(14))#

Explanation:

First, add #11# to both sides of the equation

#x^2-10x+11=-11+11#

#x^2-10x+11 = 0#

There are no nice integers which multiply together to make #+11# and add together to make #-10#. So factoring will involve using either the quadratic equation or completing the square.

Let's choose completing the square. Rewrite the above equation with spaces to prepare for the next step.

#(x^2-10x+" ")+11 + " "=0#

Take the coefficient of #x#, in this case #-10#, divide it by #2# and square the final result.

#((-10)/2)^2=25#

In the spaces provided, first add #25#, then subtract it, as follows:

#(x^2-10x+25)+11-25=0#

This makes a perfect square inside the parenthesis

#(x-5)^2+11-25=0#

#(x-5)^2-14=0#

#(x-5)^2=14#

Take the square root of both sides

#x-5=+-sqrt(14)#

#x=5+-sqrt(14)#

This gives two results

#x=5+sqrt(14)# and #x=5-sqrt(14)#

Solving both for zero gives your factors

#x-5-sqrt(14)=0# and #x-5+sqrt(14)=0#

Thus, the original equation factors as follows:

#x^2-10x+11=(x-5-sqrt(14))(x-5+sqrt(14))#

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