What is the solution set for #x^2 - 5x + 6 = 0#?

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Answer 1 · 2015-08-31T00:34:01

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

For a general form quadratic equation

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

you can determine its roots by using the quadratic formula

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

In your case, #a = 1#, #b = -5#, and #c = 6#. This means that you have

#x_(1,2) = (-(-5) +- sqrt((-5)^2 - 4 * 1 * 6))/(2 * 1)#

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

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

The two roots will thus be

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