How do you solve #9x^2=25# using the quadratic formula?

1 Answer
Jan 16, 2017

Answer:

See entire solution process below:

Explanation:

First, we need to transform this into quadratic form by subtracting #color(red)(25)# from each side of the equation to keep the equation balanced while equating to #0#:

#9x^2 - color(red)(25) = 25 - color(red)(25)#

#9x^2 - 25 = 0#

This is a special form of the quadratic equation which has the solution:

#color(red)(a)x^2 - color(blue)(b) = (sqrt(color(red)(a))x + sqrt(color(blue)(b)))(sqrt(color(red)(a))x - sqrt(color(blue)(b))) = 0#

Substituting from our quadratic gives:

#color(red)(9)x^2 - color(blue)(25) = (sqrt(color(red)(9))x + sqrt(color(blue)(25)))(sqrt(color(red)(9))x - sqrt(color(blue)(25))) = 0#

#(3x + 5)(3x - 5) = 0#

or, because the #sqrt(9) = +-3#

#(-3x + 5)(-3x - 5) = 0#

Now, we can solve each term for #0#:

Solution 1)

#3x + 5 = 0#

#3x + 5 - 5 = 0 - 5#

#3x + 0 = -5#

#3x = -5#

#(3x)/3 = -5/3#

#x = -5/3#

Solution 2)

#3x - 5 = 0#

#3x - 5 + 5 = 0 + 5#

#3x + 0 = 5#

#3x = 5#

#(3x)/3 = 5/3#

#x = 5/3#

Solution 3)

#-3x + 5 = 0#

#-3x + 5 - 5 = 0 - 5#

#-3x + 0 = -5#

#-3x = -5#

#(-3x)/(-3) = (-5)/-3#

#x = 5/3#

Solution 4)

#-3x - 5 = 0#

#-3x - 5 + 5 = 0 + 5#

#-3x + 0 = 5#

#-3x = 5#

#(-3x)/-3 = 5/-3#

#x = -5/3#

#x = 5/3# or #x = -5/3#