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

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

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Answer 1 · 2016-07-20T17:01:19

$x= 7.916 " or " 0.0842$

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..

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

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