Solving Quadratic Equation: Explaining Why "a" Must be Equal to 1

[FrogPad]

Umm... I guess I don't remember this from algebra, but I have a rather basic question.

Let's say you have the expression,
$$f(x) = 4x^2 + 5x + 1$$

And we want to find the roots.

So,

$$\frac{-5 \pm \sqrt{5-4(4)}}{2(4)}= \{ \frac{1}{2},-1 \}$$

Now, if we plug these into $f(x)$ we have:
$$f(1/2) = 9/2$$
$$f(-1) = 0$$

So obviously the quadratic equation does not yield the roots. But if we let $a=1$.

$$4(x^2+\frac{5}{4}x+\frac{1}{4})$$
Again, using the quadratic:
$$x = \{ -1,-\frac{1}{4} \}$$

So now plugging this into $f(x)$

$$f(x) = 4(x^2+\frac{5}{4}x+\frac{1}{4})$$

yields:
$$f(-1) = 0$$
$$f(\frac{-1}{4}) = 0$$

So what gives. I don't think I was ever taught that "a" has to be equal to 1 for it to hold.

If someone could clarify why this is that would be awesome. I seem to remember something about completing the square, but the thing is, is I have an engineering test on Thursday, and the last thing I want to be using my time on is figuring out why this is. But I'm curious

Thanks

 

[mbrmbrg]

Let's say you have the expression,
$$f(x) = 4x^2 + 5x + 1$$

And we want to find the roots.

So,

$$\frac{-5 \pm \sqrt{-4(4)}}{2(4)}= \{ \frac{1}{2},-1 \}$$

Recheck your quadratic equation. It should be:
$$\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$

You forgot to take the sqrt of b^2 - 4ac

 

[Gagle The Terrible]

You can also try to complete the square or find the numbers m, n such as m*n = c and m+n = b when a = 1 .

 

[FrogPad]

Recheck your quadratic equation. It should be:
$$\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$

You forgot to take the sqrt of b^2 - 4ac

Ok, cool... so I'm not crazy. It works fine for the reals. Now when I get complex conjugates, it seems to be a different story. Please tell me I'm making a simple mistake again !


$$f(s) = 10s^2 + 10s + 40$$
$$\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-10 \pm \sqrt{100-4(10)(40)}}{2(10)}=\frac{-10 \pm\sqrt{-1500}}{2(10)} = \ldots$$
$$\ldots = \left\{ \frac{-1}{2} + \frac{\sqrt{15}}{2}j \,\, , \,\, \frac{-1}{2} - \frac{\sqrt{15}}{2}j \right\} = \{ \lambda_1 \,\, , \,\, \lambda_2 \}$$
$$f(\lambda_1) = f(\lambda_2) = 0$$
So that's cool. But what about,

$$(s-\lambda_1)(s-\lambda_2)=s^2+s+4$$

If the roots were real then wouldn't that equal the original function. Maybe I'm doing drugs, because I can't believe I'm getting caught up with this right now

 

[shmoe]

It works fine for complex roots.

In your examples of s^2+s+4 and 10s^2+10s+40, remember if u and v are roots of a polynomial ax^2+bx+c, then

ax^2+bx+c=a*(x-u)*(x-v)

it's not just a product over factors corresponding to the roots, you need to take care of the leading coefficient as well.

 

[FrogPad]

ax^2+bx+c=a*(x-u)*(x-v)

it's not just a product over factors corresponding to the roots, you need to take care of the leading coefficient as well.

Ahh, that's what I was missing! Thanks man. I can't believe I forgot that