Erroneous Root

[turophile]

I got an extra root in solving a problem about a minimum distance, and I'm not sure why.

Here's the problem:

Find the point on the x-axis from which the sum of the distances to (0, 4) and (4, 2) is a minimum.

Here's what I did:

Let x be the point on the x-axis, and let s be the sum of the distances. Then:

s = sqrt[(0 – x)[sup:30tulfvf]2[/sup:30tulfvf] + (4 – 0)[sup:30tulfvf]2[/sup:30tulfvf]] + sqrt(4 – x)[sup:30tulfvf]2[/sup:30tulfvf] + (2 – 0)[sup:30tulfvf]2[/sup:30tulfvf]] = sqrt[x[sup:30tulfvf]2[/sup:30tulfvf] + 16] + sqrt[16 – 8x + x[sup:30tulfvf]2[/sup:30tulfvf] + 4] = sqrt[x[sup:30tulfvf]2[/sup:30tulfvf] + 16] + sqrt[x[sup:30tulfvf]2[/sup:30tulfvf] – 8x + 20]. We want to find x when s is a minimum.

s' = 1/2 ? 1/sqrt[x[sup:30tulfvf]2[/sup:30tulfvf] + 16] ? 2x + 1/2 ? 1/sqrt[x[sup:30tulfvf]2[/sup:30tulfvf] – 8x + 20] ? (2x – 8) = x/sqrt[x[sup:30tulfvf]2[/sup:30tulfvf] + 16] + (x – 4)/sqrt[x[sup:30tulfvf]2[/sup:30tulfvf] – 8x + 20] ? 0

? x/sqrt[x[sup:30tulfvf]2[/sup:30tulfvf] + 16] = – (x – 4)/sqrt[x[sup:30tulfvf]2[/sup:30tulfvf] – 8x + 20]
? x ? sqrt[x[sup:30tulfvf]2[/sup:30tulfvf] – 8x + 20] = (– 1) ? (x – 4) ? sqrt[x[sup:30tulfvf]2[/sup:30tulfvf] + 16]
? x[sup:30tulfvf]2[/sup:30tulfvf] ? (x[sup:30tulfvf]2[/sup:30tulfvf] – 8x + 20) = (x – 4)[sup:30tulfvf]2[/sup:30tulfvf] ? (x[sup:30tulfvf]2[/sup:30tulfvf] + 16)
? x[sup:30tulfvf]4[/sup:30tulfvf] – 8x[sup:30tulfvf]3[/sup:30tulfvf] + 20x[sup:30tulfvf]2[/sup:30tulfvf] = (x[sup:30tulfvf]2[/sup:30tulfvf] – 8x + 16) ? (x[sup:30tulfvf]2[/sup:30tulfvf] + 16) = x[sup:30tulfvf]4[/sup:30tulfvf] + 16x[sup:30tulfvf]2[/sup:30tulfvf] – 8x[sup:30tulfvf]3[/sup:30tulfvf] – 128x + 16x[sup:30tulfvf]2[/sup:30tulfvf] + 256
? – 12x[sup:30tulfvf]2[/sup:30tulfvf] + 128x – 256 = 0
? 12x[sup:30tulfvf]2[/sup:30tulfvf] – 128x + 256 = 0
? 3x[sup:30tulfvf]2[/sup:30tulfvf] – 32x + 64 = 0
? x = {– (– 32) ± sqrt[(– 32)[sup:30tulfvf]2[/sup:30tulfvf] – 4(3)(64)]}/[2(3)] = (32 ± sqrt[1024 – 768])/6 = (32 ± sqrt[256])/6 = (32 ± 16)/6 = 16/6 or 48/6 = 8/3 or 8

Now my textbook (and my graph of s) indicates that s is a minimum when x = 8/3. Any my graph of s doesn't show any evidence of a maximum or a minimum at x = 8. So the root I got of x = 8 is an error. Where did it come from?

 

[galactus]

That is not necessarily an error but what is known as an extraneous root. It occurs, but we do not need it.

The 8 is just another root of the quadratic you solved. All quadratic s have two roots, though both may not be necessary.

 

[soroban]

Hello, turophile!

I got an extra root in solving a problem about a minimum distance, and I'm not sure why.

Here's the problem: Find the point on the x-axis from which the sum of the distances to (0, 4) and (4, 2) is a minimum.

Here's what I did:

Let \(\displaystyle x\) be the point on the x-axis, and let \(\displaystyle s\) be the sum of the distances. Then:

\(\displaystyle s \:=\: \sqrt{(x - 0)^2 + (4 - 0)^2} +\sqrt{(x-4)^2 + (2 - 0)^2} = \sqrt{x^2 + 16} + \sqrt{x^2 - 8x + 20}\)

We want to find \(\displaystyle x\) when \(\displaystyle s\) is a minimum.

\(\displaystyle s' \;=\;\frac{1}{2}\cdot\frac{2x}{\sqrt{x^2+16}} + \frac{1}{2}\cdot\frac{2x-4}{\sqrt{x^2-8x+20}} \;=\;0\) .[1]

. . . . \(\displaystyle \frac{x}{\sqrt{x^2 + 16}} \;=\; - \frac{x - 4}{\sqrt{x^2 - 8x + 20}}\)

. . . . \(\displaystyle x\sqrt{x^2 - 8x + 20} \;=\; -(x - 4)\sqrt{x^2 + 16}\)

\(\displaystyle \text{Square both sides: }\;x^2(x^2 - 8x + 20) \;=\;(x - 4)^2(x^2 + 16)\)

. . . . \(\displaystyle x^4 - 8x^3 + 20x^2 \;=\; (x^2 - 8x + 16)(x^2 + 16)\)

. . . . \(\displaystyle x^4 - 8x^3 + 20x^2 \;=\;x^4 + 16x^2 - 8x^3 - 128x + 16x^2 + 256\)

. . . . \(\displaystyle 12x^2 - 128x + 256 \;=\; 0\)

. . . . \(\displaystyle 3x^2 - 32x + 64 \;=\; 0\) .[2]

. . . . \(\displaystyle (3x-8)(x-8) \;=\;0\)

. . . . \(\displaystyle x \;=\;\tfrac{8}{3},\:8\)


Now my textbook (and my graph of \(\displaystyle s\)) indicates that \(\displaystyle s\) is a minimum when \(\displaystyle x = \tfrac{8}{3}\)

And my graph of \(\displaystyle s\) doesn't show any evidence of a maximum or a minimum at \(\displaystyle x = 8\)

So the root I got of \(\displaystyle x = 8\) is an error. . Well, not exactly.
Where did it come from?

First of all, your work is absolutely correct . . . Nice going!

Note that both roots work in the quaratic equation [2].

But \(\displaystyle x = 8\) does not work in the original equation [1]

Recall that you squared both sides of the equation.
. . And this can create "extraneous" roots.

So, as galactus pointed out: \(\displaystyle x = 8\) is not an error.
It is an extra (unneeded) root of that quadratic equation.