How do you solve #4(x+1)^2=100#?

1 Answer
Aug 1, 2016

Answer:

#x = 4" "# or #" "x = -6#

Explanation:

The key to this problem is the fact that for a given number #a in RR#, you have

#a^2 = a * a" "# and #" "a^2 = (-a) * (-a)#

so right from the start you know that you're looking for two values of #x#, since for #a in RR#, you have

#color(purple)(|bar(ul(color(white)(a/a)color(black)(sqrt(a^2) = |a| = {(+ a, " for "a >=0), (-a, " for "a < 0):})color(white)(a/a)|)))#

The first thing to do here is divide both sides of the equation by #4#

#(color(red)(cancel(color(black)(4))) * (x+1)^2)/color(red)(cancel(color(black)(4))) = 100/4#

#(x+1)^2 = 25#

If you take the square root of both sides of the equation

#sqrt((x+1)^2) = sqrt(25)#

you will end up with

#(x+1) = + 5" "# or #" "(x+1) = -5#

This will get you

#x+1 = 5 implies x= 4" "# or #" " x+1 = -5 implies x= -6#

Therefore, the original equation has two possible solutions

#x = 4" "# or #" "x = -6#

Do a quick check to make sure that the calculations are correct

#x = 4:" " 4 * (4 + 1)^2 = 100#

#4 * 5^2 = 100" "color(green)(sqrt())#

#x = -6:" " 4 * (-6 + 1)^2 = 100#

#4 * (-5)^2 = 100" "color(green)(sqrt())#