How do you solve #x^2+15x=0#?

2 Answers
Alan P.
Jul 15, 2016

#color(blue)(x=0)# or #color(blue)(x=-15)#

Explanation:

Factoring #x^2+15x=0#
gives
#color(white)("XX")x(x+15)=0#

which implies
either
#color(white)("XX")x=0#
or
#color(white)("XX")(x+15=0)#
#color(white)("XXXXX")rarr x=-15#

EZ as pi
Jul 15, 2016

#x= 0, or x = -15#

Explanation:

The first approach is always to try and factorise the quadratic. Although this is not a trinomial, there is a common factor.

#x(x+15)=0#

Because the product is 0, one of the two factors must be equal to 0.

Either #x = 0, or x+ 15 =0," in which case " x = -15#

This gives two solutions, exactly what we expect with a quadratic.

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