How do you solve #-2x^2 - 7x + 4=0# using completing the square?

1 Answer
Jun 16, 2015

Answer:

The answers are #x=1/2# and #x=-4#.

Explanation:

First, factor the coefficient of #x^2# out of the first two terms to get #-2x^2-7x+4=-2(x^2+7/2 x)+4=0#.

Next, take the coefficient of #x# inside the parentheses, #7/2#, divide it by 2 to get #7/4#, and then square that number to get #49/16#. Add this number inside the parentheses and then "balance" it by adding #-2*49/16# on the other side of the equation to get #-2(x^2+7/2 x+ 49/16)+4=-2*49/16=-49/8#.

The reason this trick is a good idea is that the expression #x^2+7/2 x+ 49/16# is a perfect square. It equals #(x+7/4)^2#, so the equation becomes #-2(x+7/4)^2+4=-49/8#, which is equivalent to #-2(x+7/4)^2=-81/8# and #(x+7/4)^2=81/16#.

Now take the #\pm# square root of both sides to get #x+7/4=\pm 9/4#, leading to two solutions #x=9/4-7/4=2/4=1/2# and #x=-9/4-7/4=-16/4=-4#.

You should check these in the original equation:

#x=1/2\Rightarrow -2(1/2)^2-7(1/2)+4=-1/2-7/2+4=-4+4=0#

#x=-4\Rightarrow -2(-4)^2-7(-4)+4=-32+28+4=0#