How do you solve #2x^2+5x=0# using the quadratic formula?

1 Answer
Jan 29, 2017

Answer:

Express in the form #ax^2+bx+c=0# then apply the formula to find:

#x = 0" "# or #" "x = -5/2#

Explanation:

Given:

#2x^2+5x=0#

we can express this equation as:

#2x^2+5x+0 = 0#

This is in the form:

#ax^2+bx+c = 0#

with #a=2#, #b=5# and #c=0#

It has roots given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-5+-sqrt((-5)^2-4(2)(0)))/(2*2)#

#color(white)(x) = (-5+-sqrt(25))/4#

#color(white)(x) = (-5+-5)/4#

So one root is:

#x = (-5+5)/4 = 0#

and the other is:

#x = (-5-5)/4 = -10/4 = -5/2#

#color(white)()#
Footnote

You really would not normally choose to use the quadratic formula in this example.

Instead note that both terms are divisible by #x#, so it factors like this:

#0 = 2x^2+5x = x(2x+5)#

Which has solutions #x=0# and #x=-5/2#