How do you solve #9x^2-4=0#?

2 Answers
Oct 21, 2017

Answer:

#x=\pm\frac{2}{3}#

Explanation:

#9x^2-4=0#

Move the #4# to the RHS

#9x^2=4#

Divide both sides by #9#

#x^2=\frac{4}{9}#

Take the square root of both sides

#\sqrt{x^2}=\pm\sqrt{\frac{4}{9}}#

Simplify

#x=\pm\frac{2}{3}#

#\thereforex=-\frac{2}{3},\qquad\qquad\qquad x=\frac{2}{3}#

Oct 21, 2017

Answer:

#x= +-sqrt(4/9#

In decimal #x = +-0.67#

Explanation:

#9x^2 - 4=0#
Let start by adding #color(red)(4)# to both sides
#9x^2 cancel(- 4) cancelcolor(red)(+4)=0+color(red)(4)#
#9x^2 = 4#
Divide both sides by #color(green)(9)#
#(cancel9x^2)/cancelcolor(green)(9) = 4/color(green)(9)#
#x^2 = 4/9#
We need to take square root #sqrt#
#x= +-sqrt(4/9)#