How do you solve #25x ^ { 2} + 6= 9#?

1 Answer
Jul 3, 2017

See a solution process below:

Explanation:

First, subtract #color(red)(6)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#25x^2 + 6 - color(red)(6) = 9 - color(red)(6)#

#25x^2 + 0 = 3#

#25x^2 = 3#

Next, divide each side of the equation by #color(red)(25)# to isolate #x^2# while keeping the equation balanced:

#(25x^2)/color(red)(25) = 3/color(red)(25)#

#(color(red)(cancel(color(black)(25)))x^2)/cancel(color(red)(25)) = 3/25#

#x^2 = 3/25#

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

#sqrt(x^2) = +-sqrt(3/25)#

#x = +-sqrt(3/25)#

Now, use this rule of radicals to simplify the result:

#sqrt(color(red)(a)/color(blue)(b)) = sqrt(color(red)(a))/sqrt(color(blue)(b))#

#x = +-sqrt(color(red)(3)/color(blue)(25))#

#x = +-sqrt(color(red)(3))/sqrt(color(blue)(25))#

#x = +-sqrt(color(red)(3))/5#