How do you solve #(sqrt2)x² - x - 3(sqrt2) =0#?

1 Answer
Stefan V.
Aug 31, 2015

#x_(1,2) = (1 +- 5)/(2sqrt(2))#

Explanation:

You know that for a general form quadratic equation

#color(blue)(ax^2 + bx + c = 0)#

you can find its roots by using the quadratic formula

#color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a))#

In your case, #a = sqrt(2)#, #b = -1#, and #c = -3sqrt(2)#. This means that you have

#x_(1,2) = ((-1) +- sqrt((-1)^2 - 4 * sqrt(2) * (-3sqrt(2))))/(2 * sqrt(2))#

#x_(1,2) = (1 +- sqrt(1 + 12 * 2))/(2sqrt(2))#

#x_(1,2) = (1 +- 5)/(2sqrt(2))#

The two solutions to this equation will be

#x_1 = (1 + 5)/(2sqrt(2)) = color(green)((3sqrt(2))/2)" "# and #" "x_2 = (1 - 5)/(2sqrt(2)) = color(green)(-sqrt(2))#

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