How do you solve #x^2+8x=33# by completing the square?

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Answer 1 · 2017-06-11T01:18:34

#x_1= 3#

#x_2=-11#

#x^2+8x=33#

How to complete the square:

If #x^2+bx+c = 0# , then #x^2+bx+c -= (x+b/2)^2-(b/2)^2 + c#

#x^2+8x=(x+4)^2-16=33#

#(x+4)^2=49#

#x+4=+-7#

#x=-4+-7#

#x_1=3#

#x_2=-11#

Answer 2 · 2017-06-11T01:34:12

#x=3, x=-11#

Set up the problem like this:

#x^2+8x+ ("a number")=33#

To complete the square, we divide #8# by #2# and then square it.

#(8/2)^2=16#

We now have:

#x^2+8x+16=33+16#

(We add 16 to both sides to keep the equation balanced)

Simplify by factoring the left side and adding up the right side:

#(x+4)^2=49#

Solve for #x# by applying the square root property:

#sqrt((x+4)^2)=sqrt49#

#x+4=+-7#

#x=-4+-7#

#x=3, x=-11#