How do you solve #2x^2-5=0#?

1 Answer
Aug 21, 2015

Answer:

#x_(1,2) = +- sqrt(10)/2#

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 #1 = sqrt(2)/sqrt(2)#

#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)#