Quadratic Equations: where does the formula come from?

By Brad Rodgers (P1930) on Wednesday, March 1, 2000 - 08:17 pm :

I have two questions about quadratic equations.

First, how is the result

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

Also, given that y=ax² +bx+c on the cartesian plane, how do we know that the vertex's x coordinate is -(b/2a)?

Any help would be much appreciated.

By Sean Hartnoll (Sah40) on Wednesday, March 1, 2000 - 09:25 pm :

First question:

We start with $a x^{2} + b x + c = 0$ , to be solved for $x$ .

'Complete the square' (this is a standard think to do), to get:

$a {(x + \frac{b}{2 a})}^{2} - \frac{b^{2}}{4 a} + c = 0$

Multiply this out if you don't see how it comes from the previous line).

Moving things to one side and taking square root:

$x + \frac{b}{2 a} = \sqrt{\frac{b^{2}}{4 a^{2}} - \frac{c}{a}}$

Rearranging:

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

Note there is a $\pm$ (meaning $+$ or $-$ ). So if the square root exists and is nonzero then there are two solutions. This is what you might expect because a parabola can intersect the $x$ -axis at two points, provided the vertex is low enough.

To find the vertex coordinate, you need to use calculus. A theorem of Calculus is that you find the minimum or maximum points of a curve (in this case the vertex) by what is called differentiating and then setting the result of this equal to zero. In this case we differentiate

$y = a x^{2} + b x + c$

to get

$y' = 2 a x + b$

Setting equal to zero: $2 a x + b = 0$

So $x = - (b / 2 a)$

Sean

By Brad Rodgers (P1930) on Wednesday, March 1, 2000 - 10:19 pm :

Thanks

Brad

By James Lingard (Jchl2) on Wednesday, March 1, 2000 - 10:24 pm :

Brad,

You can also answer your second question without using calculus in the following way.

You start by 'completing the square' as Dan did for your first question, to get the equation in the form:

$a {(x + \frac{b}{2 a})}^{2} + c - \frac{b^{2}}{4 a} = 0$

I'll assume that $a > 0$ , and so the vertex of the parabola is just the point where $y$ has its minimum value. (If $a < 0$ then it's exactly the same idea, just you're looking for the maximum value).

Now, the second half of the right hand side of the equation (i.e. the $(c - b^{2} / 4 a)$ bit) is just a constant, and is independent of the value of $x$ . So $y$ has its minimum value at the point where the first half of the right hand side of the equation has its minimum value.

But, just as $x^{2}$ is never negative, no matter what value $x$ is and so $x^{2}$ has its minimum value when $x = 0$ , we can do the same with $(x + b / 2 a)^{2}$ . This is always at least 0, and has its minimum value (i.e. 0) not when $x = 0$ but when $x + b / 2 a = 0$ . This shows you that $x = - b / 2 a$ .

Also, you can use the completing the square method to quickly show you what the minimum/maximum value is - i.e. the $y$ -coordinate of the vertex. Can you see how to do this?

Hope that helps - as always, please feel free to ask if there's anything you don't understand.

James.

By Sean Hartnoll (Sah40) on Wednesday, March 1, 2000 - 11:02 pm :

Sorry for giving a calculus proof, when of course there is a simple proof which doesn't require calculus, as James gave above.

I'll try to briefly explain what differentiation is and why the derivative is zero at a maximum or minimum.

Suppose we have a curve y = f(x).

Fix a point on the x-axis, say x₀ . We can define

y₀ = f(x₀ )

Now suppose we increase x₀ by a small amount dx, this means y₀ will also change by a small amount, say dy (see picture). So

y₀ + dy = f(x₀ +dx)

But y₀ = f(x₀ ), so rearranging

dy = f(x₀ +dx) - f(x₀ )

This is how much y₀ increases when we increase x₀ by dx. We can consider the ratio of the increase in dy and dx:

dy/dx = [f(x₀ +dx)-f(x₀ )]/dx

Now we let dx tend towards zero (i.e. get arbitrarily small) and we call the limit f'(x₀ ), called the derivative of f at x₀ .

We can see this value will be related to the slope of the curve at the point (x₀ ,y₀ ). The more the curve is sloped, the larger dy will be for given dx, and so the larger dy/dx will be.

If we are at an extreme point (maximum of minimum) then dy=0! This is because for a very small increase in x, dx, y does not change at all because the curve is horizontal near that point. This is why f'(x₀ ) = 0 at the vertex of a parabola.

To complete the proof, I should give you a way of calculating derivatives, there are some general formulae for different functions, but it is easy to derive the ones we need for a parabola

i) f(x) = c (constant) then f'(x) = 0

because f'(x) = limit (f(x+dx)-f(x))/dx
= limit (c-c)/dx
= 0

ii) f(x) = bx then f'(x) = b

because f'(x) = limit (f(x+dx)-f(x))/dx
= limit (bx + bdx - bx)/dx
= limit bdx/dx = b

iii) f(x) = ax² then f'(x) = 2ax

because f'(x) = limit (a(x+dx)² - ax² )/dx
= limit (2axdx + adx² )/dx
= limit (2ax + adx)
= 2ax

the last term vanishes in the limit as dx goes to 0.

Hope this is intelligible and that it clarifies the proof I gave earlier,

Sean
Diagram

By Brad Rodgers (P1930) on Thursday, March 2, 2000 - 04:17 pm :

How do you 'complete the square'. I understand what you do when doing this, but I don't see how you can derive this result from the original equation given.

Thanks,

Brad

By Sean Hartnoll (Sah40) on Thursday, March 2, 2000 - 05:33 pm :

The idea of completing the square is to make an expression look 'nicer' and be easier to deal with. It is important to have in mind always the expression:

(m+n)² = m² + 2mn + n² (*)

Now given the expression

x² + bx

we try to make it look as much like the right hand side of (*) as possible so we set x = m and bx = 2mn.

This implies bm = 2mn, so n = b/2.

Finally, to look like (*), we need the term n² = b² /4 which we don't have. So we add and substract it from our expression, giving:

x² + bx + b² /4 - b² /4

Using (*) this equals

(x+b/2)² - b² /4

And we have completed the square.

The expression in the quadratic ax² + bx + c is slightly more complicated because it has an 'a' and a 'c' in it. Perhaps you can see how to deal with that?

Hope this helps,

Sean

By Brad Rodgers (P1930) on Friday, March 3, 2000 - 02:29 am :

I think I understand this and think I see how you deal with a.

Thanks,

Brad.