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

2 Answers
Apr 17, 2016

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

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