What is x^2-8x-20=0 solving by completing the square?

2 Answers
Oct 9, 2017

#x=10#

Explanation:

#x^2-8x-20 = 0 #

Add 20 to both sides...

#x^2-8x = 20 #

When completed we should have a function of the form #(x+a)^2#. This function expanded would be #x^2+2ax+a^2 #. If #2ax=-8x#, then #a=-4#, meaning our term will be #(x-4)^2#. Expanded this would give us #x^2-8x+16#, so to complete the square we have to add 16 to both sides...

#x^2-8x+16 = 20+16 #

Now change it into our #(x+a)^2# form...

# (x-4)^2 = 36 #

Square root both sides:

#x-4 = 6#

And finally add 4 to both sides to isolate x.

#x=10#

Oct 9, 2017

#x=10,\qquad\qquad x=-2#

Explanation:

First, move the #c# value to the RHS:

#x^2-8x=20#

Add #(\frac{b}{2})^2# to both sides:

#x^2-8x+(\frac{-8}{2})^2=20+(\frac{-8}{2})^2#

Simplifying the fractions:

#x^2-8x+16=20+16#

Now that the LHS is a perfect square, we can factor it as #(x-\frac{b}{2})^2#

#(x-4)^2=36#

Taking the real (non-principal) square root:

#\sqrt{(x-4)^2}=\sqrt{36}#

Simplifying:

#x-4=\pm 6#

Isolating for #x#:

#x=\pm 6+4#

#\quad x=-6+4,\qquad x=6+4#

#\therefore x=-2,\qquad\qquad x=10#