How do you solve # -3x^2 - 9x = 0#?

1 Answer
Jun 25, 2016

Answer:

#x = -3# and #x = 0#

Explanation:

There are two ways to do this:

First Way: Factoring

Now this one is quite nice as we can just remove a common factor from both of them... they are both divisible by #-x#:

#-x(3x + 9) = 0#

Now we need to make this equation #0#, therefore #-x# or #3x +9# must equal #0# as any number multiplied by #0 = 0#...

So we know one solution for #x# must therefore be #0# (as #-(0) * x = 0#, #x# being any number)

we also know that #3x + 9 = 0#, so we can solve this:

#3x + 9 = 0 -># subtract 9

#3x = -9 -># divide by 3

#x = -3#

So the two solutions for #x# are #0# and #-3#

Second Way: Quadratic Formula

This way is a bit harder and provides unneccesary difficulty for this question, though shows how the formula can be applied on any quadratic equation:

We know for any equation in the form:

#ax^2 + bx + c = 0#

#x# can be found by the equation:

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

Now in the equation it is less obvious to see as there is no c value but:

#a = -3 -># as #-3x^2#

#b = -9 -># as #-9x#

#c = 0 -># as not specified

Therefore substituting in:

#x=(-(-9)+- sqrt((-9)^2 -4(-3)(0)))/(2(-3))#

#x=(9+- sqrt(81))/(-6)#

#x =(9+-9)/(-6)#

#x=(18)/-6 = -3# and #x=(0)/-6=0#

So the two solutions for #x# are #0# and #-3#

Checking

We can test these solutions by adding them back in:

#-3(0)^2 - 9(0) = 0#

#0 - 0 = 0#

and

#-3(-3)^2 - 9(-3) = 0#

#(-3)(9) + 27 = 0#

#-27 + 27 = 0#

Therefore we know our answers are right.