How do you solve #2x ^ { 2} - 25x = 0#?

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Answer 1 · 2017-05-13T12:49:18

#x_1 = 12.5, x_2 = 0#

#2x^2 - 25x = 0#

Solve with the quadratic formula

For quadratic equation of the form #ax^2+bx+c=0# the solutions are

#x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}#

For#" " a=2, b=-25, c=0 #

#x_{1,2}=\frac{-\ (-25\ )\pm \sqrt{\ (-25\ )^2-4\cdot \ 2\ *0}}{2\ xx\2}#

#x_{1,2}=\frac{\ (25 \ )\pm \sqrt{\ (-25\ )^2- 0}}{2\ xx\2}#

#x_{1,2}=\frac{\ (25 \ )\pm \ \ (-25\ )}{2\ xx\2}#

#x_{1,2}=\frac{\ (25 \ )\pm \ \ (25\ )}{2\ xx\2}#

#x_{1,2}=\frac{\ (25 \ )\pm \ \ (25\ )}{4}#

#x_1 = (25 + 25)/4 = 50/4 = 25/2 = 12.5#

#x_2 = (25-25)/4 = 0/4 = 0#

Answer 2 · 2017-05-13T14:22:42

#x =0 " or "x = 12.5#

The usual method with quadratic equations is to factorise first if possible. If not you have to resort to another method.

#2x^2 -25x =0" "larr# there is a common factor.

#x(2x-25)=0#

We now have two factors, set each equal to #0#.

#x = 0" or " 2x-25 =0#

#x =0 " or " x = 25/2 = 12.5#