mathmath333
  • mathmath333
question
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmath333
  • mathmath333
any short way to solve this . \(\large \color{black}{\begin{align} \left|\dfrac{x^2-5x+4}{x^2-4}\right|\leq 1\hspace{.33em}\\~\\ \end{align}}\)
ganeshie8
  • ganeshie8
\(\large \left|\dfrac{x^2-5x+4}{x^2-4}\right| = \left|\dfrac{(x^2-4)-5x+8}{x^2-4}\right| = \left|1+\dfrac{-5x+8}{x^2-4}\right|\) For this to be \(\le 1\), clearly we must have \[-2\lt\dfrac{-5x+8}{x^2-4}\lt 0\]
mathmath333
  • mathmath333
u mean \(\large \color{black}{\begin{align} -2\leq\dfrac{-5x+8}{x^2-4}\leq 0\hspace{.33em}\\~\\ \end{align}}\) ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

mathmath333
  • mathmath333
i am also confused on how to further solve this . \(\large \color{black}{\begin{align} -2\leq\dfrac{-5x+8}{x^2-4}\leq 0\hspace{.33em}\\~\\ \end{align}}\)
ganeshie8
  • ganeshie8
Ahh yes, consider two cases : Case1 : \(x^2-4\gt 0\) multiply \(x^2-4\) through out and get \[-2(x^2-4 )\le -5x+8 \le 0 \\~\\\implies -2x^2 \le -5x \le -8 \\~\\\implies 2x^2\ge 5x\ge 8 \\~\\\implies x\ge \frac{5}{2} \land x\ge \frac{8}{5}\\~\\\implies x\ge \frac{5}{2} \]
ganeshie8
  • ganeshie8
similarly you can work the other case i don't like this method yeah too boring
UsukiDoll
  • UsukiDoll
\[\large \color{black}{\begin{align} \left|\dfrac{x^2-5x+4}{x^2-4}\right|\leq 1\hspace{.33em}\\~\\ \end{align}}\] Since we have the absolute value, we need to consider two cases. One when we have an all positive problem like this \[\frac{x^2-5x+4}{x^2-4} \leq 1 \] or \[-(\frac{x^2-5x+4}{x^2-4}) \leq 1 \] I'm going to try the all positive case \[\frac{x^2-5x+4}{x^2-4} \leq 1 \] so I would distribute the \[x^2-4 \] to the other side \[x^2-5x+4 \leq 1(x^2-4)\] \[x^2-5x+4 \leq x^2-4 \] \[x^2-5x+4 -x^2+4 \leq 0\] (and then put everything on the left... combine like terms) \[x^2-x^2-5x+4 +4 \leq 0\] \[-5x+8 \leq 0\] (simplify and divide. since we have a negative number, the sign switches from \[\leq \] to \[\geq \] \[-5x\leq -8\] \[x\geq \frac{8}{5}\]
UsukiDoll
  • UsukiDoll
there has got to be a way to make Latex nicer. this is ugly :P
mathmath333
  • mathmath333
the above needs to consider this two cases also \(\dfrac{x^2-5x+4}{x^2-4}\geq 0,\ \ \dfrac{x^2-5x+4}{x^2-4}< 0\)
UsukiDoll
  • UsukiDoll
I said I just picked one case :P I'm aware that there are two.
mathmath333
  • mathmath333
is there a way to transform the question in to this. \(\large \color{black}{\begin{align} \left|\dfrac{x^2-5x+4}{x^2-4}+\color{black}{constant}\right|\leq \color{red}{0}\hspace{.33em}\\~\\ \end{align}}\)
UsukiDoll
  • UsukiDoll
maybe subtract that 1 on both sides? But I have never done it, so I don't know if it's a legal math move.
mathmath333
  • mathmath333
i m not saying substract \(1\) lol
UsukiDoll
  • UsukiDoll
\[\frac{-x^2+5x-4}{x^2-4} \leq 1\] \[-x^2+5x-4 \leq 1(x^2-4)\] \[-x^2+5x-4 \leq x^2-4\] \[-x^2-x^2+5x-4+4 \leq 0\] \[-2x^2+5x\leq 0\] \[x(-2x+5)\leq 0\] x would be 0... blah don't want that case \[-2x+5 \leq 0 \] \[-2x \leq -5 \] \[x\geq \frac{5}{2} \] Looks like there is no faster way of doing this. You just have to use the absolute value rules which is consider the cases where we have an all positive situation and one where the negative sign is being distributed.
anonymous
  • anonymous
Why you have chosen 1 to subtract @UsukiDoll ?
UsukiDoll
  • UsukiDoll
We have two cases thanks to the absolute value remember we either have all positive or we have a negative sign attached to the problem and have to distribute it and then do the same process as above. @ganeshie8 had the same answers I have.
UsukiDoll
  • UsukiDoll
\(\color{blue}{\text{Originally Posted by}}\) @waterineyes Why you have chosen 1 to subtract @UsukiDoll ? \(\color{blue}{\text{End of Quote}}\) didn't do it.. just in case that move didn't make sense :P
anonymous
  • anonymous
Which move?

Looking for something else?

Not the answer you are looking for? Search for more explanations.