How do you solve using the completing the square method #2x^2-4x-14=0#?

2 Answers
Apr 28, 2018

Answer:

The two solutions are #x=+-2sqrt2+1#.

Explanation:

First, divide the whole equation by two:

#2x^2-4x-14=0#

#x^2-2x-7=0#

Then, move the constant to the other side:

#x^2-2x=7#

Next, identify the square binomial on the left side. We know that #(x-1)^2# expands to #x^2-2x+1#, so if we add #1# to both sides, we can use this backwards:

#x^2-2xcolor(red)+color(red)1=7color(red)+color(red)1#

#(x-1)^2=8#

Now, square root both sides:

#x-1=+-sqrt8#

#x-1=+-2sqrt2#

#x=+-2sqrt2+1#

Those are the solutions. Hope this helped!

Apr 28, 2018

Answer:

#x=1pm2sqrt2#

Explanation:

Factor out 2:

#2[x^2-2x]-14# #lArr# notice we didn't factor the constant.

Complete the square of the brackets:

#2[(x-1)^2-1]-14#

Multiply the brackets:

#2(x-1)^2-2-14#

Simplify:

#2(x-1)^2-16#

We can always check this:

#(x-1)^2=x^2-2x+1#

#2(x^2-2x+1)=2x^2-4x+2-16 -> 2x^2-4x-14#

Solving:

#2(x-1)^2-16=0#

#2(x-1)^2=16#

#(x-1)^2=8#

#x-1=pmsqrt8#

#x=1pmsqrt8#

#-> x=1pm2sqrt2#