# Math Help - Another inverse function issue...

1. ## Another inverse function issue...

Given that f : x -> x^2 + 6x, and g: x -> 2x - 1 ... where x is a real number.

Find the two values of x such that fg(x) = gf(x), giving each answer in the form $p + q \sqrt{3}$

I have tried solving it by equating by solving it for both of the functions but that's not working. How should I proceed with it?

2. Hello,
Originally Posted by struck
Given that f : x -> x^2 + 6x, and g: x -> 2x - 1 ... where x is a real number.

Find the two values of x such that fg(x) = gf(x), giving each answer in the form $p + q \sqrt{3}$
Well...

$f(g(x))=(2x-1)^2+6(2x-1)=\text{expand everything}=4x^2+8x-5$

$g(f(x))=2(x^2+6x)-1=\dots=2x^2+12x-1$

Now solve for x in :
$4x^2+8x-5=2x^2+12x-1$

that is $2x^2-4x-4=0$

$x^2-2x-2=0$

in order to do that, find the discriminant, and use your head... ^^

3. Sorry, but I should have posted the steps. I just wanted to know if I had to right idea. Ok, so far my steps were correct but at the last step, I was finding this using quadratic formula.

$\frac {2 \pm \sqrt{4 + 8}}{2}$

$\frac {2 \pm \sqrt{3} \sqrt {4}}{2}$

$\frac {2 \pm 2 \sqrt{3}}{2}$

The answer I am supposed to get is $1 \pm \sqrt{3}$ but I am getting $2 \pm \sqrt {3}$. I had, similarly, tried it without simplification at the last step ... can't seem to figure out what's wrong.

4. Originally Posted by struck
Sorry, but I should have posted the steps. I just wanted to know if I had to right idea. Ok, so far my steps were correct but at the last step, I was finding this using quadratic formula.

$\frac {2 \pm \sqrt{4 + 8}}{2}$

$\frac {2 \pm \sqrt{3} \sqrt {4}}{2}$

$\frac {2 \pm 2 \sqrt{3}}{2}$

The answer I am supposed to get is $1 \pm \sqrt{3}$ but I am getting $2 \pm \sqrt {3}$. I had, similarly, tried it without simplification at the last step ... can't seem to figure out what's wrong.
Yeah, by discriminant, I meant what you did.
And yes, it would have been better to show the steps avoiding some useless lines

Anyway, you've got the answer !

$\frac{2\pm 2 \sqrt{3}}{2} \neq 2 \pm \sqrt{3}$
I don't know how you got that !

$\frac{2 \pm 2 \sqrt{3}}{2}=\frac 22 \pm \frac{2 \sqrt{3}}{2}=1 \pm \sqrt{3}$

5. OMG. My brain wasn't working well I guess. Thanks.