# Thread: Basic "solve for x" problem

1. ## Basic "solve for x" problem

"If $f(x) = x^2 - 3x$, determine the values of $x$ for which $f(x - 3) \leq 0$"

This is for only two marks, but I'm not quite sure how to go about doing this... Any tips would be great!

2. ## Re: Basic "solve for x" problem

$f(x-3) = (x-3)^2 - 3(x-3) = (x-3)^2 - 3x +9$

Can you find the values of X for which the above expression is $\leq 0$

hint: Write it as a quadratic then find the solutions that give exactly 0. Then plot the graph and the range of values is clear

Spoiler:

http://www.wolframalpha.com/input/?i=%28x-3%29^2+-+3x+%2B9+%3C%3D+0

3. ## Re: Basic "solve for x" problem

Got it, thank you.

4. ## Re: Basic "solve for x" problem

Originally Posted by SpringFan25
$f(x-3) = (x-3)^2 - 3(x-3) = (x-3)^2 - 3x +9$

Can you find the values of X for which the above expression is $\leq 0$

hint: Write it as a quadratic then find the solutions that give exactly 0. Then plot the graph and the range of values is clear

Spoiler:

http://www.wolframalpha.com/input/?i=%28x-3%29^2+-+3x+%2B9+%3C%3D+0
Or if you didn't want to waste time graphing, completing the square solves that quadratic inequality beautifully...

\displaystyle \begin{align*}(x - 3)^2 - 3x + 9 &\leq 0 \\ x^2 - 6x + 9 - 3x + 9 &\leq 0 \\ x^2 - 9x + 18 &\leq 0 \\ x^2 - 9x + \left(-\frac{9}{2}\right)^2 - \left(-\frac{9}{2}\right)^2 + 18 &\leq 0 \\ \left(x - \frac{9}{2}\right)^2 - \frac{81}{4} + \frac{72}{4} &\leq 0 \\ \left(x - \frac{9}{2}\right)^2 - \frac{9}{4} &\leq 0 \\ \left(x - \frac{9}{2}\right)^2 &\leq \frac{9}{4} \\ \left|x - \frac{9}{2}\right| &\leq \frac{3}{2} \\ -\frac{3}{2} \leq x - \frac{9}{2} &\leq \frac{3}{2} \\ 3 \leq x &\leq 6 \end{align*}

5. ## Re: Basic "solve for x" problem

whoops... disregard. I misinterpreted the original question... My bad