How do you solve #5x^2 - 90 = 0#?

1 Answer
EZ as pi
Jul 12, 2016

#x = +3sqrt2" or " x = -3sqrt2#

Explanation:

Usually to solve a quadratic equation, we make it equal to 0.
However in this case, as there is no term in x, we can use a different method.

#5x^2 = 90 " "÷ 5#

#x^2 = 18 " find the" +- sqrt#

#x = +-sqrt18#

#x = +3sqrt2" or " x = -3sqrt2#

If the constant had been a perfect square we could have factorised as difference of two squares, but 18 is not a square.

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