Quadratic Formula False?

Question

The quadratic formula seems to find intercepts correctly but not the right graph. What am I missing?

For example, if $0=2x^2+7x+5$, then it can be solved with the quadratic formula to give us $x=-5/2$ or $-1$. When I graph $(x+5/2)(x+1)$ it looks like this.

However, when I graph $y=2x^2+7x+5$ it looks like this.

These graphs are clearly different, but I don't know why. However, if I use PSF or cross method I get the correct factorization, but I don't want to use them because they don't always work.

Answer 1

When you expand $(x+5/2)(x+1)$, you get $$x^2 + x + \frac{5}{2} x + \frac{5}{2} = x^2 + \frac{7}{2} x + \frac{5}{2} = \frac{1}{2}(2x^2 + 7x + 5).$$

That's why the two graphs look different: one is half of the other.

Solving the equation $0 = ax^2 + bx + c$ using the quadratic formula does not mean that $ax^2 + bx + c = (x-r_1)(x-r_2)$ where $r_1, r_2$ are the roots. It means $$ax^2 + bx + c = a(x - r_1)(x - r_2).$$ You should review the proof of the quadratic formula, which uses the technique of completing the square.

Answer 2

You forgot to multiply by $2$. If the roots of a polynomial are $r_1,\ldots, r_n$ then the polynomial is a multiple of $(x-r_1)(x-r_2)\cdots(x-r_n)$.

Answer 3

The quadratic formula tells you the zeros, but two functions can have the same zeroes while being different overall.

In particular, you're conflating your original polynomial $$2x^2+7x+5$$ and the polynomial $$(x+{5\over 2})\cdot (x+1).$$ Notice that when we multiply the latter out we get $$x^2+{7\over 2}x+{5\over 2},$$ which is a multiple of the original. Of course, if $f$ is a multiple of $g$ then $f$ and $g$ have the same roots (since $a\cdot 0=0$ regardless of what $a$ is). One important takeaway is that knowing the roots of a polynomial does not tell you everything about it.