How do you solve #32x^2-18=0#?

3 Answers
Apr 26, 2017

Answer:

See the solution process below:

Explanation:

First, add #color(red)(18)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#32x^2 - 18 + color(red)(18) = 0 + color(red)(18)#

#32x^2 - 0 = 18#

#32x^2 = 18#

Next, divide each side of the equation by #color(red)(32)# to isolate the #x^2# term while keeping the equation balanced:

#(32x^2)/color(red)(32) = 18/color(red)(32)#

#(color(red)(cancel(color(black)(32)))x^2)/cancel(color(red)(32)) = (2 xx 9)/color(red)(2 xx 16)#

#x^2 = (color(red)(cancel(color(black)(2))) xx 9)/color(red)(color(black)(cancel(color(red)(2))) xx 16)#

#x^2 = 9/16#

Now, take the square root of each side of the equation to solve for #x# while keeping the equation balanced. Remember, taking the square root of a number produces a negative and positive result:

#sqrt(x^2) = sqrt(9/16)#

#x = sqrt(9)/sqrt(16)#

#x = +-3/4#

Apr 26, 2017

Answer:

x = 0.75

Explanation:

#32x^2-18=0#
#32x^2-18+18=0+18#
#32x^2=18#
#32x^2/32=18/32#
#x^2=0.5625#
#sqrt(x^2)=sqrt(0.5625)#
#x=0.75#

Apr 26, 2017

Answer:

#color(blue)(x=3/4# or #color(blue)(x=-3/4#

Explanation:

#32x^2-18=0#

Take out the common factor first

#:.2(16x^2-9)=0#

#:.2(4^2x^2-3^2)=0#

Divide both sides by # 2#

#:.(4^2x^2-3^2)=0#

#:.(4x-3)(4x+3)=0#

#:.4x-3=0,4x+3=0#

#:.4x=3,4x=-3#

#:.color(blue)(x=3/4,x=-3/4#

substitute #color(blue)(x=3/4=0.75#

#:.32(color(blue)0.75)^2-18=0#

#:.32(0.5625)-18=0#

#18-18=0#

substitute #x=-3/4=-0.75#

#:.32(color(blue)-0.75)^2-18=0#

#:.32(0.5625)-18=0#

#:.18-18=0#