How do you solve the equation 2(x+2)^2=72?

4 Answers
Mar 22, 2017

"answer "{x=-8" , "x=4}

Explanation:

2(x+2)^2=72

"divide both sides by 2"

(cancel(2)(x+2)^2)/cancel(2)=(72)/2

(x+2)^2=36

"take square root both sides"

(x+2)=+-6

"if "x+2=-6" , then " x=-8

"if "x+2=6 " , then " x=4

Mar 22, 2017

x = 4 or -8

Explanation:

2(x+2)^2 = 72

divide by 2:

(x+2)^2 = 36

square root:

x+2 = 6 or -6

subtract 2:

x = 4 or -8

Mar 22, 2017

x={4,-8}

Explanation:

First divide both sides by two to get (x+2)^2=36. Next take the square root of both sides to get x+2=+-6. Now split this into two equations for both the positive and negative six. These look like: x+2=6 and x+2=-6. For both equations, subtract two from both sides which gives you x=4 and x=-8. Hence x={4,-8}.

Mar 22, 2017

x=-8" or " x=4

Explanation:

Divide both sides by 2

rArr2/2(x+2)^2=72/2

rArr(x+2)^2=36

Take the color(blue)"square root of both sides"

sqrt((x+2)^2)=color(red)(+-)sqrt36

rArrx+2=color(red)(+-)6

• x+2=color(red)(6)

subtract 2 from both sides.

xcancel(+2)cancel(-2)=6-2

rArrx=4

• x+2=color(red)(-6)

rArrx=-8

color(blue)"as a check"

Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.

x=4to2(4+2)^2=2xx36=72to" true"

x=-8to2(-8+2)^2=2xx36=72to" true"

rArrx=-8" or " x=4" are the solutions"