How do you solve the quadratic equation by completing the square: #x(6x-5) = 6#?

1 Answer
Jul 22, 2015

Answer:

The two solutions are #x=3/2# and #x=-2/3#.

Explanation:

First use the distributive property to write #x(6x-5)=6# as #6x^2-5x=6#. Next factor the coefficient of #x^2# out on the left side to write it as #6(x^2-(5/6)x)=6#.

We can now complete the square: Take the coefficient of #x# inside the parentheses, the #-5/6#, divide it by 2 to get #-5/12#, then square that number to get #25/144#.

Next, add the #25/144# inside the parentheses on the left and compensate for that (balance the equation) by adding #6*25/144=25/24# to the right hand side:

#6(x^2-(5/6)x+25/144)=6+25/24=(144+25)/24=169/24#.

The reason this is a good idea is that this technique has made the expression inside the parentheses on the left a perfect square. This last equation can be written as

#6(x-5/12)^2=169/24# (note also the appearance of the #-5/12# again...this is no coincidence).

To finish, divide both sides by #6# to get #(x-5/12)^2=169/144# and then take the square root of both sides, allowing a #\pm# sign on the right to get #x-5/12=\pm 13/12#.

Now add #5/12# to everything to get #x=5/12 \pm 13/12#. The two solutions are #x=18/12=3/2# and #x=-8/12=-2/3#.