How do you solve #2x^2-5=0#?
1 Answer
Aug 21, 2015
Explanation:
You need to take three steps in order to solve this equation
- add
#5# to both sides of the equation
#2x^2 - color(red)(cancel(color(black)(5))) + color(red)(cancel(color(black)(5))) = 0 + 5#
#2x^2 = 5#
- divide both sides of the equation by
#2#
#(color(red)(cancel(color(black)(2)))x^2)/color(red)(cancel(color(black)(2))) = 5/2#
#x^2 = 5/2#
- take the square root of both sides to solve for
#x#
#sqrt(x^2) = sqrt(5/2)#
#x_(1,2) = +- sqrt(5)/sqrt(2)#
You can simplify this further by rationalizing the denominator if the fraction. To do that, multiply the fraction by
#x_(1,2) = +- (sqrt(5) * sqrt(2))/(sqrt(2) * sqrt(2))#
#x_(1,2) = +- sqrt(10)/2#
This means that your equation has two solutions,
#x_1 = color(green)(sqrt(10)/2)" "# or#" "x_2 = color(green)(-sqrt(10)/2)#
