How do you solve #(p + 7) ^ { 2} = 216#?

3 Answers
Dec 7, 2017

#p=-7+-6sqrt6#

Explanation:

#color(blue)"take the square root of both sides"#

#sqrt((p+7)^2)=+-sqrt216larrcolor(blue)"note plus or minus"#

#rArrp+7=+-6sqrt6#

#"subtract 7 from both sides"#

#pcancel(+7)cancel(-7)=-7+-6sqrt6#

#rArrp=-7+-6sqrt6larrcolor(red)"exact solutions"#

Dec 7, 2017

#p=-7+-6sqrt(6)#

Explanation:

#(p+7)^2=216#

#rarr p+7 = +-sqrt(216)#

#color(white)("XXX")226 =2^2 * 3^2 * 6 =6^2 * 6#

#rarr p+7 =+-6sqrt(6)#

#rarr p=-7+-sqrt(6)#

Dec 7, 2017

#p ~~ 7.70#

Explanation:

Alright, so we need to isolate the #p# on one side of the equation so that we can solve this.

To do that, we are going to start with taking the square root of both sides.

#sqrt((p+7)^2) = +-sqrt(216)#

The square root and the square will cancel out so all we have to do is finish the equation!

#p+7~~14.70#
#p +7-7~~ 14.70-7#
#p ~~ 7.70#

Also, I would like to add that is not the exact number, IT IS AN APPROXIMATION. If you would like to have the exact number, see below. Since #sqrt(216)# can not be completely squared, it would have to be reduced to

#p+7=+-sqrt(216)#
#p+7=+-6sqrt(6)#
#p=-7+-6sqrt(6)#

Hope that helps!
~Chandler Dowd