How do you solve #4x ^ { 2} + 3= 24#?

1 Answer
Feb 5, 2017

See the entire solution process below:

Explanation:

Step 1) Subtract #color(red)(3)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#4x^2 + 3 - color(red)(3) = 24 - color(red)(3)#

#4x^2 + 0 = 21#

#4x^2 = 21#

Step 2) Divide each side of the equation by #color(red)(4)# to isolate #x^2# while keeping the equation balanced:

#(4x^2)/color(red)(4) = 21/color(red)(4)#

#(color(red)(cancel(color(black)(4)))x^2)/cancel(color(red)(4)) = 5.25#

#x^2 = 5.25#

Step 3) Take the square root of each side of the equation to solve for #x# while keeping the equation balanced. When taking the square root the answer will be for both the negative and positive root:

#sqrt(x^2) = +-sqrt(5.25)#

#x = +-sqrt(5.25) = +-2.291# rounded to the nearest thousandth.