Q5 - Equations with an additonal restricted variable

[ExamFever]

Q5 - Equations with an "additonal" restricted variable

Homework Statement

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.

Homework Equations

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?

 

[Mentallic]

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?

 

[eumyang]

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.

Given the basic quadratic equation
$$ax^2 + bx + c = 0$$

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

 

[Mentallic]

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 $$(-b/a)^2-2c/a$$

 

[ExamFever]

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...

 

[Mentallic]

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.

 

[ExamFever]

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...

 

[Mentallic]

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.