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

Question

Nathan Axel · Accepted Answer

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

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#. These are things to look for 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#

Hazel Anglin · Answer

undefined To solve the quadratic equation -2x^2 - 7x + 4 = 0 using completing the square:

1. Move the constant term to the other side: -2x^2 - 7x = -4.
2. Divide the entire equation by the coefficient of x^2, which is -2: x^2 + (7/2)x = -2.
3. Add the square of half the coefficient of x (7/4)^2 to both sides: x^2 + (7/2)x + (7/4)^2 = -2 + (7/4)^2.
4. Factor the left side into a perfect square trinomial: (x + 7/4)^2 = -2 + 49/16.
5. Simplify the right side: (x + 7/4)^2 = -32/16 + 49/16 = 17/16.
6. Take the square root of both sides: x + 7/4 = ±√(17/16).
7. Solve for x: x = -7/4 ± √(17/16).
8. Simplify the square root: x = -7/4 ± (1/4)√17.
9. Final solutions: x = (-7 ± √17)/4.