How do you solve #x^2+14x=0 # using the quadratic formula?

2 Answers
Apr 17, 2016

Answer:

#x = -14, 0#

Explanation:

Quadratic Formula
#x = (-(b) pm sqrt{(b)^2 - 4ac})/{2a}#

Substitute:
a term = 1
b term = 14
c term = 0

#x = (-(14) pm sqrt{(14)^2 - 4(1)(0)})/{2(1)}#

Evaluate:
#x = (-14 pm sqrt{196 - 0})/2#
#x = (-14 pm sqrt{196})/2#
#x = (-14 pm 14)/2#

Solve to find the two roots of x
#x = (-14 + 14)/2 = 0/2 = 0#
#x = (-14 - 14)/2 = (-28)/2 = -14#

#x = -14, 0#

Apr 17, 2016

Answer:

Check my full answer, please.

Explanation:

Not disagreeing with Dèv, but there is another(and even easier) way to solve it:

In a quadratic function, whenever #c=0# its possible to solve it this way. First, you put #x# in evidence and then you find 2 possibilities:
#x^2+14x = 0#
#x(x+14)=0#

So, the 2 possibilities: #x = 0# or #x+14 = 0#. The first possibility is already solved, because if #x=0#:
#0*(0+14) = 0*(14) = 0#. As we can see, its correct.

The second possibility we may finish solving it:
#x+14 = 0 =># The #14# is positive and goes to the other side of the equality as #-14#, but #0-14 = -14#, so #x = -14#.

Now, we do solve it the same way we did it before:
#-14*(-14+14) = -14*(0) = 0#

So, the roots of a quadratic function with #c = 0# are:
#ax+b = 0# and #x = 0#