# Q5 - Equations with an additonal restricted variable

1. Aug 26, 2010

### ExamFever

Q5 - Equations with an "additonal" restricted variable

1. The problem statement, all variables and given/known data

P1
Determine the parameter m such that the sum of the squares of the roots of equation

x2 - (m-2)x + m - 3 = 0

is minimal.

P2
Determine the values of parameter p, so that the solution of equation

p - (1/x) = (p2 - 2x - 5)/((p+3)x)

is greather than 1.

2. Relevant equations

3. The attempt at a solution

I would like to know whether there is any specific way of calculating equations with an additonal "restricted" variable, like in the two problems above.

Is a rational reasoning followed by a trial and error used or is more specific calculation possible?

P1

x2 - (m-2)x + m - 3 = 0

The sum of the squares must be minimal, thus:

(ax - b)(cx - d) where ax2 + cx2 must be minimal

roots are minimal when

(x - 1)2 or (x + 1)2 namely 12 or (-1)2 which equals 1

This form

x2 - 2x +1
x2 +2x +1

Can be achieved when

x2 - (4-2)x + 4 - 3 = 0

Thus m = 4

Are my calculations here correct? Is this the correct way to solve this problem?

P2

p - (1/x) = (p2 - 2x - 5)/((p+3)x)

I find this problem a lot more difficult. Reasoning here is a lot trickier (for me at least). I can only see that p can not equal -3

When either working the equation out of fractions, ie

(px - 1)(p +3) = p2 - 2x - 5

or moving everything to one side, I am still left with one equation and "two" unknowns. Another approach I tried when anaylizing this problem was simply putting in a value for x, x=1, x=2 and x=3 all return values of p=-2

How should I approach such a problem?

2. Aug 26, 2010

### Mentallic

Re: Q5 - Equations with an "additonal" restricted variable

P1)
Well you know that for a quadratic $$ax^2+bx+c=0$$ then sum of the roots are -b/a, so all you have is m-2 for the sum of the roots. But we want the sum of these roots to be minimalistic, which means as close to zero as possible so we let m-2=0 thus m=2. Substituting m=2 into the quadratic gives us x2-1=0 and if you graph this, the roots are 1 and -1, and the sum of these is 0, so that's what we wanted.
Your answer of (x-1)2=0 has two roots both at 1, and the sum of these is 2 which is not minimalistic.

You can also do this a longer way, but you get the same answer and you might find it interesting. If we use the quadratic formula to find the roots of x in terms of m, so we have

$$x=\frac{m-2\pm\sqrt{(m-2)^2-4(m-3)}}{2}$$

The discriminant $$\Delta = m^2-4m+4-4m+12=m^2-8m+16=(m-4)^2$$

So we can now simplify this into

$$x=\frac{m-2\pm(m-4)}{2}$$

And so the two roots are (lets take the positive of the $\pm$ and then the negative)

$$x_1=m-3$$

$$x_2=1$$

This is quite interesting, no matter what value of m we use, we always have a root of 1. Anyway, so the sum of these is m-3+1=m-2 which is what we got before.

P2)
In the same way we used the quadratic formula above to find x in terms of m, we want x in terms of p.
Can you do this?

3. Aug 27, 2010

### eumyang

Re: Q5 - Equations with an "additonal" restricted variable

Unless I'm misreading this, the OP said to find m such that the sum of the squares of the roots of the equation
$$x^2 - (m - 2)x + m - 3 = 0$$
that is minimal.

$$ax^2 + bx + c = 0$$

the sum of the squares of the roots is
$$\left( \frac{-b + \sqrt{b^2 - 4ac}}{2a} \right)^2 + \left( \frac{-b - \sqrt{b^2 - 4ac}}{2a} \right)^2$$

This can be simplified to something relatively simple. After doing so, plug in
a = 1, b = -(m - 2), c = m - 3

and you'll get a quadratic on m. The graph of this quadratic on m will open upward, so the vertex will be the minimum point. Put this quadratic in vertex form and you'll get your answer. I'm getting an integer answer here.

69

4. Aug 27, 2010

### Mentallic

Re: Q5 - Equations with an "additonal" restricted variable

Oh yeah sorry I missed that.

Again it can be solved both ways.

Let the two roots be $\alpha$ and [itex]\beta[/tex]

$$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$$

So we have $$(-b/a)^2-2c/a$$

5. Aug 28, 2010

### ExamFever

Re: Q5 - Equations with an "additonal" restricted variable

Ok, thank you both for the replies.

If I continue from

$$\left( \frac{-b + \sqrt{b^2 - 4ac}}{2a} \right)^2 + \left( \frac{-b - \sqrt{b^2 - 4ac}}{2a} \right)^2$$
I simplify into

(b2-2ac)/a2

Putting in the values of a, b and c:

-(m-2)2 - 2(1 * (m-3)) divided by 12 thus the fraction disappears
putting in the values of m we get:

-(m-2)(m-2) - 2 (m-3)
-m2 -4m -4 - 2m + 6 Is this correct or should it have been (-(m-2))2?
-m2 - 6m + 2

-m2 thus concave downward...from your answer I get it should be concave upward thus (-(m-2))2?

m2 - 6m + 10

Vertex form:

y-1 = (m-3)2

Vertex at (3,1) thus m = 3?

P2
I get the idea described above but I can not get this equation into standard parabola form ax2 + bx + c. I have the feeling this might be a line?

p2x-p2+3px+2x-p+2=0

Is the equation when I put everything to one side. However, I do not see how this could be a variation of the standard form of either a line or parabola...

Last edited: Aug 28, 2010
6. Aug 28, 2010

### Mentallic

Re: Q5 - Equations with an "additonal" restricted variable

P1) What you've done seems correct. I just want to point out that we were searching for an m such at the sum of the squares of the parabola is minimalistic, which means closest to zero from either end as possible. What the parabola in terms of m tells us is that, say, for m=3 the sum of the squares is 1, for m=4 the sum is 2 etc. And since at m=3 this is the closest we get to 0, that is the answer.
But if the parabola was something different like y+1=(m-3)2 this equation has real roots so the roots would be the answer, not the vertex.

P2) All you have to do is get x in terms of p. Don't worry about whether it is a line or a parabola or whatnot, knowing this information isn't going to help you answer the question.

7. Aug 29, 2010

### ExamFever

Re: Q5 - Equations with an "additonal" restricted variable

Mentallic, thanks for the reply. I realize getting x in terms of p should be easy but somehow I keep getting stuck.

p - (1/x) = (p2 - 2x -5)/((p+3)x)

I've tried to first remove fractions:

(px-1)(p+3) = p2 - 2x - 5
(px-1)(p+3)-p2 = -2x - 5

However, here I'm left with a px term on the LHS which I cannot seem to get rid off. Everything I do (the above being my best try I think) leaves a term of px which I cannot get rid off, making it impossible to express x in a terms of p with no x's remaining:

x = - 0.5 (p2x - p2 + 3 px - p + 2)

I do not see how I can go from here...

8. Aug 29, 2010

### Mentallic

Re: Q5 - Equations with an "additonal" restricted variable

Those p's are getting to you :tongue:

If I asked you to isolate for x in the following, I bet you can do it.

ax+bx+c+d=0

The same idea applies to your problem.