How do you solve #x^2+10x+16=0# by completing the squares?

2 Answers
Jul 17, 2018

Answer:

#x = -8, -2#

Explanation:

Given: #x^2 + 10x + 16 = 0#

To complete the square put the two #x# terms on the left and the constant on the right of the equation:

#color(red)(x^2 + 10x) = -16#

Complete the square by multiplying the #x#-term by #1/2#:
# 1/2 * 10 = 5#

and adding the square of this number to the right side of the equation: #5^2 = 25#

#(x +5)^2 = -16 color(blue)(+ 25)#

UNDERSTANDING CHECK:
#(x+5)^2 = color(red)(x^2 + 10x) color(blue)(+ 25).#

The #+25# was not in the original equation. If we add #+25# to one side of the equation, we must add the same amount to the other side of the equation to keep it balanced.

#(x+5)^2 = 9#

To solve, square root both sides of the equation:

#sqrt((x+5)^2) = +- sqrt(9)#

#x + 5 = +- 3#

# x = -5 +- 3#

#x = -5 + 3 = -2, " "x = -5 -3 = -8#

#x = -8, -2#

Jul 17, 2018

Answer:

#x = -2 or -8#

Explanation:

The process of completing the square is done by adding a missing term to an expression so as to create the square of a binomial.

In #x^2 +10x+16=0," "16# is not the required constant, so move it to the right side.

#x^2 +10x" "=-16#

The required constant is determined from #color(blue)((b/2)^2)# where #b=10#

#color(blue)((10/2)^2 = 5^2 =25)#

Add this to both sides of the equation:

#x^2 +10xcolor(blue)(+25)=-16 color(blue)(+25)#

The left side is now the square of a binomial, ie a perfect square.

#" "(x+5)^2 = 9" "larr# find the square root of both sides.

#" "x+5 = +-sqrt9 = +-3#

#" "x = +-3-5#

This leads to two solutions:

#x = +3-5 = -2" "or" "x = -3-5=-8#