How do you solve using the completing the square method #x^2 + 8x + 16 = 36#?

1 Answer

#x=-10,2#

Explanation:

By completing the square first we will make is #=0#, and we do this by #-36#. this gets us:

#x^2+8x-20# as #16-36=-20#

Then as we complete the square we want to half the middle term and place it in brackets and square it, and put the end term on or off (depending on if a #+# or #-#):

#x^2+8x-20 -> (x+4)^2 -20#

Remember, we always minus the squared number in the bracket. #4^2=16# therefore we #-16#

#-> (x+4)^2-16-20#

By collecting like terms:

#(x+4)^2-36.#

This is in complete the square form, so now we solve.

First set it to #0#:

#(x+4)^2-36=0#

Get rid of the extra value on the end, and by doing this we do cancel out the 36, so we do the opposite, so add 36. Remember to do this to both sides.

#(x+4)^2=36#

We do not want a squared bracket, therefore we do the opposite which is square root both sides

#-> (x+4)^2=36 -> (x+4)=pmsqrt36#

Always remember the #pm# when square rooting.

As you may be able to see, using our squared numbers we know that #sqrt36# can be simplified.

#sqrt36=6# as #6xx6=36#

Therefore we can simplify this to:

#x+4=pm6# and we can solve like a usual equation.

We want to get rid of the #4#, therefore, we do the opposite to cancel it out, which is to #-4#, doing this to both sides.

#x=-4pm6#

#x=-4-6=-10#

or

#x=-4+6=2#

This cannot be simplified further, so, therefore, this is our final answer.