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find $g(x)$

$f(g(x)) = h(x)$

$f(x) = x^{2}-6x+2$

$h(x) = 9x^{2}-24x+9$

My work:

$g(x)^{2}-6(g(x))+2 = h(x)$

$g(x)^{2}-6(g(x))-(9x^{2}-24x+7)$

And then what's next step?please provide explaination as well Thank You!!

3+- sqr(9-(-9x^2-24x+7))

3+- sqr(9+9x^2+24x-7)

Then what do i do?

  • 1
    You're very close! Your last line should include "$=0$", since you essentially subtracted $h(x)$ from both sides. Then just view this as a quadratic equation, where $g(x)$ is the variable to solve for (regard $x$ as a constant during this process if it helps), so your quadratic is of the form $$G^2-6G-C = 0$$ – 2017-01-12

2 Answers 2

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Note $$\eqalign{f(x)&=(x-3)^2-7\\h(x)&=(3x-4)^2-7}$$ It follows that $$f(3x-1)=h(x).$$

Hope this helps.

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Hint:

Let $y=g(x)$,

$$y^2-6y+2 = 9x^2-24x+9$$

$$y^2-6y-9x^2+24x-7=0$$ This is a quadratic equation in $y$, solve for $y$

$$y=\frac{6\pm \sqrt{6^2-4(-9x^2+24x-7)}}{2}$$