Given is the function: f(x)= (x^2-4x) / (x^2+2)

For which p ( element R ) contains the set of solutions of the equation f (x) = p exactly one element ?
To give the found p the set of solutions.

How do I solve this task correctly? Look, I can fill in for p, p=-1 , p=1 and p=2 .

So I find the answers= p=-1 with x=1
p=1 with with x=1/2
p=2 with x=-2

But is this the right way? I filled these values for p , because I saw this in the answer book.
If I did not have the answer book, how do I know which values of p should I choose?

2. ## Re: Question about comparison

$\dfrac{x^2-4x}{x^2+2}-p = 0$

$\dfrac{x^2-4x}{x^2+2} - \dfrac{p(x^2+2)}{x^2+2} = 0$

$\dfrac{x^2(1-p)-4x-2p}{x^2+2}=0$

Rational expression equals zero when the numerator equals zero ...

$x=\dfrac{4 \pm \sqrt{(-4)^2-4(1-p)(-2p)}}{2(1-p)}$

to get a single solution for $x$, the discriminant must equal zero. Can you finish?

3. ## Re: Question about comparison

Equivalently:
$\frac{x^2- 4x}{x^2+ 2}= p$
$x^2- 4x= px^2+ 2p$
$(1- p)x^2- 4x= 2p$
$x^2- \frac{4}{1- p}x= \frac{2p}{1- p}$

Complete the square:
$x^2- \frac{4}{1- p)x+ \frac{4}{(1- p)^2= \frac{2p}{1- p}+ \frac{4}{(1- p)^2= \frac{2p- 2p^2+ 4}{(1- p)^2}$
$(x- \frac{2}{1- p})^2= -2\frac{p^2- p- 2}{(1- p)^2}$

Solve by taking the square root of both sides. If the right side is negative, there is no root. If the right side is positive we can take both "+" and "-" and there are two real roots. If the right side is 0, there is only one root.

4. ## Re: Question about comparison

Originally Posted by skeeter
$\dfrac{x^2-4x}{x^2+2}-p = 0$

$\dfrac{x^2-4x}{x^2+2} - \dfrac{p(x^2+2)}{x^2+2} = 0$

$\dfrac{x^2(1-p)-4x-2p}{x^2+2}=0$

Rational expression equals zero when the numerator equals zero ...

$x=\dfrac{4 \pm \sqrt{(-4)^2-4(1-p)(-2p)}}{2(1-p)}$

to get a single solution for $x$, the discriminant must equal zero. Can you finish?

If I calculate for the discriminant of this, I find: p=-1 and p =2 , I find two solutions, so the discriminant is not equal to zero?

5. ## Re: Question about comparison

Originally Posted by HallsofIvy
Equivalently:
$\frac{x^2- 4x}{x^2+ 2}= p$
$x^2- 4x= px^2+ 2p$
$(1- p)x^2- 4x= 2p$
$x^2- \frac{4}{1- p}x= \frac{2p}{1- p}$

Complete the square:
$x^2- \frac{4}{1- p)x+ \frac{4}{(1- p)^2= \frac{2p}{1- p}+ \frac{4}{(1- p)^2= \frac{2p- 2p^2+ 4}{(1- p)^2}$
$(x- \frac{2}{1- p})^2= -2\frac{p^2- p- 2}{(1- p)^2}$

Solve by taking the square root of both sides. If the right side is negative, there is no root. If the right side is positive we can take both "+" and "-" and there are two real roots. If the right side is 0, there is only one root.

On the left I find: x=0 .

On the right I find : p=-1 and p =2 . Is this what you meant ?

6. ## Re: Question about comparison

Originally Posted by westerwolde
Given is the function: f(x)= (x^2-4x) / (x^2+2)
For which p ( element R ) contains the set of solutions of the equation f (x) = p exactly one element ?
Originally Posted by skeeter
$\dfrac{x^2-4x}{x^2+2}-p = 0$
Rational expression equals zero when the numerator equals zero ...
$x=\dfrac{4 \pm \sqrt{(-4)^2-4(1-p)(-2p)}}{2(1-p)}$
to get a single solution for $x$, the discriminant must equal zero.
Originally Posted by westerwolde
On the left I find: x=0 .
On the right I find : p=-1 and p =2 . Is this what you meant ?
Yes those are the correct answers. Each of those values for $\bf{p}$ gives a single solution for the original question.

7. ## Re: Question about comparison

But I don't know what you mean by "x= 0". Those values of p make the part after $\pm$ equal to 0 so that the single solution to the equation is $x= \frac{4\pm 0}{2(1- p)}$. If p= -1 that is $x= \frac{4}{4}= 1$ and if p= 2 that is $x= \frac{4}{-2}= -2$.

8. ## Re: Question about comparison

Okay, this is clear. But I still do not understand how they come in p=1 with with x=1/2 in the answer book.. Or will this be a mistake in the answer book?

9. ## Re: Question about comparison

Originally Posted by westerwolde
Okay, this is clear. But I still do not understand how they come in p=1 with with x=1/2 in the answer book.. Or will this be a mistake in the answer book?
Finding the number of solutions of the equation $f(x)=p$ is equivalent to finding the number of solutions of the equation

$x^2(1-p)-4x-2p=0$

Case 1. $p=1$ this is a linear equation which has a unique solution

Case 2. $p\neq 1$ we have a quadratic equation whose discriminant must be 0 if we want to have exactly one solution