# Find the values of $k$ for which the equation $(f\circ g)(x) = x$ has two equal roots

I'm busy doing a problem which asks the above considering the following:

$$f(x) = 4x - 2k\text{ and }g(x) = 9/(2-x)$$

As far as I know roots usually refer to quadratics and even when doing the composition I still can't figure out how to find the two equal roots.

After the composition I get the following:

$$(f\circ g)(x) = (36/(2-x)) - 2k$$

Now, I don't know how I am suppose to turn that into a quadratic to find out the roots, unless the problem is just semantics and the question is really asking for the intercept and not the roots.

Any advice or guidance?

Thank you!

Bernard

• You need to set your composite expression equal to $x$ (as stated in the title), after multiplying through by $(2-x)$ the resulting equation will be quadratic in $x$. – okrzysik Feb 19 '16 at 10:45

## 1 Answer

You said that

$$(f\circ g)(x)=\frac{36}{2-x}-2k$$

But as both you and okrzysik pointed out, the composition $(f\circ g)(x)=x$ and so

$$(f\circ g)(x)=\frac{36}{2-x}-2k=x$$

So multiplying both sides by $2-x$ yields

$$(2-x)x=36-2k(2-x)\Rightarrow x^2+(2k-2)x+(36-4k)$$

So the solution by QF is

$$x=\frac{2-2k\pm\sqrt{4k^2-8k+4-4(36-4k)}}{2}$$ $$=1-k\pm\sqrt{k^2+2k-35}$$

For the problem at hand, we need the two roots to be equal, so

$$1-k+\sqrt{k^2+2k-35}=1-k-\sqrt{k^2+2k-35}$$

$$\sqrt{k^2+2k-35}=-\sqrt{k^2+2k-35}$$

$$k^2+2k-35=0 \Rightarrow (k+7)(k-5)=0...$$

and solve for $k$ from here.... But, to check to see if this is valid or not, realize that if $(f\circ g)(x)=x$, then $g=f^{-1}$ and so

$$f^{-1}(x)=\frac{x+2k}{4}$$

and since $g=f^{-1}$, we get that

$$\frac{x+2k}{4}=\frac{9}{2-x}$$

Solving by cross multiplication yields the same quadratic as above.

• I would just like to clarify, under "So the solution by QF is", should the 8k not be -8k? – Bernard Feb 20 '16 at 8:31
• Hand slapping face, yes @BernardFrankel, – Eleven-Eleven Feb 20 '16 at 12:39