# Is this true?

1. Jul 30, 2007

### lLovePhysics

If $$b^{2}-4ac=0$$ and a is not equal to 0, then the graph of $$y=ax^{2}+bx +c$$ has only one x-intercept.

I say it is true because, according to the quadratic formula, x would equal $$\frac{-b}{2a}$$.

2. Jul 30, 2007

### nicktacik

Sure. The quadratic formula tells us the X intercepts are when

$$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

So if $$b^2 - 4ac = 0$$

Then we are left with

$$x=\frac{-b}{2a}$$

Which is a unique solution.

3. Jul 30, 2007

### bongswaka

disagree

i say no.if we let a=0 the x^2 variable vanishes and we are left with a straight line.

Last edited: Jul 30, 2007
4. Jul 30, 2007

### nicktacik

Well he specified "and a is not equal to 0" in the original post