How do you solve #8x=3x^2 - 1# using completing the square?

1 Answer
Jul 28, 2015

Answer:

#x_1 = 4/3 + sqrt(19)/3#, #x_2 = 4/3 - sqrt(19)/3#

Explanation:

Start by writing your equation in quadratic form

#8x = 3x^2 - 1#

#3x^2 - 8x - 1 = 0#

To solve this equation by completing the square you first need to find a way to write the left side of the equation as a square of a binomial.

Move the term that does not contain #x# or #x^2# to the right side of the equation.

#3x^2 - 8x = 1#

Now divide everything by 3 to get

#(color(red)(cancel(color(black)(3))) * x^2)/color(red)(cancel(color(black)(3))) - 8/3x = 1/3#

#x^2 - 8/3x = 1/3#

Next, divide the coefficient of the #x#-term by 2, square it, then add the result to both sides of the equation. In your case, you have

#(-8/3) * 1/2 = -4/3#, then

#(-4/3)^2 = 16/9#

Your quadratic will become

#x^2 - 8/3x + 16/9 = 1/3 + 16/9#

The left side of the equation can be written as

#(x^2 - 8/3x + 16/9) = (x-4/3)^2#

This will get you

#(x - 4/3)^2 = 19/9#

Take the square roots of both sides to find the two solutions

#sqrt((x-4/3)^2) = sqrt(19/9)#

#x-4/3 = +- sqrt(19)/3 => x_(1,2) = 4/3 +- sqrt(19)/3#

The two solutions will be

#x_1 = color(green)(4/3 + sqrt(19)/3# and #x_2 = color(green)(4/3 - sqrt(19)/3)#