How do you solve #2(x-5)^2=3#?

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Answer 1 · 2015-08-21T10:05:43

#x = 5 +- sqrt(6)/2#

The first thing you need to do is isolate #(x-5)^2# on one side of the equation by dividing both sides by #2#.

#(color(red)(cancel(color(black)(2))) * (x-5)^2)/color(red)(cancel(color(black)(2))) = 3/2#

#(x-5)^2 = 3/2#

Now you need to take the square root of both sides

#sqrt((x-5)^2) = sqrt(3/2)#

#x-5 = +- sqrt(3)/sqrt(2)#

Rationalize the denominator by multiplying the fraction by #1 = sqrt(2)/sqrt(2)# to get

#x-5 = +- (sqrt(3) * sqrt(2))/(sqrt(2) * sqrt(2)) = +- sqrt(6)/2#

Finally, isolate #x# on one side by adding #5# to both sides of the equation

#x - color(red)(cancel(color(black)(5))) + color(red)(cancel(color(black)(5))) = 5 +- sqrt(6)/2#

#x_(1,2) = 5 +- sqrt(6)/2#

This equation will thus have two solutions,

#x_1 = color(green)(5 + sqrt(6)/2)" "# or #" "x_2 = color(green)(5 - sqrt(6)/2)#