How do you solve for #x# in #3(x-4)^2 - 6 = 40# ?

1 Answer
Jul 20, 2016

Answer:

#x= 7.916 " or " 0.0842#

Explanation:

This is a specific form of a quadratic which does not have an #x# term. We can do this in a similar way to the method of completing the square, because the square is already there.

Isolate the bracket

#3(x-4)^2 - 6 = 40#

#3(x-4)^2 = 46#

#(x-4)^2 = 46/3" "# Find the square root of each side and solve .

#x-4 = +-sqrt(46/3)#

#x = sqrt(46/3) +4 " or " x = -sqrt(46/3)+4#

#x= 7.916 " or " 0.0842#

You could also multiply the whole expression out and then solve it as quadratic in the usual way with the formula.
This way is quicker..