mathmath333
  • mathmath333
solve for \(x\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} x^2-|5x+8|>0\hspace{.33em}\\~\\ \end{align}}\)
misty1212
  • misty1212
HI!!
misty1212
  • misty1212
work in cases solve if \(x<-\frac{8}{5}\) and if \(x>-\frac{8}{5}\)

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mathmath333
  • mathmath333
i worked m onfised
misty1212
  • misty1212
??
mathmath333
  • mathmath333
*confused
misty1212
  • misty1212
if \(x>-\frac{8}{5}\) then \[|5x+8|=5x+8\] right ?
misty1212
  • misty1212
if it is not clear how i got that say so and i will explain
mathmath333
  • mathmath333
i dk how u got it
misty1212
  • misty1212
the absolute value is a piece wise function
misty1212
  • misty1212
let me write it as one hold on
mathmath333
  • mathmath333
ok i got it
misty1212
  • misty1212
ok so if \(5x+8>0\) then \(|5x+8|=5x+8\) and \(5x+8>0\) if \(x>-\frac{8}{5}\) so case 1:\(x>-\frac{8}{5}\) you solve \[x^2-(5x+8)>0\] or \[x^2-5x-8>0\]
misty1212
  • misty1212
once you solve that, the actual answer is the intersection of your solution and \(x>-f\rac{8}{5}\) then repeat with \(x<-\frac{8}{5}\) solving \[x^2+5x+8>0\]
misty1212
  • misty1212
ach
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} x>\dfrac{5+\sqrt{57}}{2},\ x<\dfrac{5-\sqrt{57}}{2} \hspace{.33em}\\~\\ \end{align}}\)
misty1212
  • misty1212
intersection of your solution and \(x>-\frac{8}{5}\)
misty1212
  • misty1212
looks good to me
mathmath333
  • mathmath333
lol m still confused
mathmath333
  • mathmath333
this is solution of \(x^2-5x-8>0\) \(\large \color{black}{\begin{align} x>\dfrac{5+\sqrt{57}}{2},\ x<\dfrac{5-\sqrt{57}}{2} \hspace{.33em}\\~\\ \end{align}}\)
mathmath333
  • mathmath333
@phi
mathmath333
  • mathmath333
for \(x^2+5x+8>0\) \(\large \color{black}{\begin{align} -\infty
phi
  • phi
You began by assuming x >= -8/5 = -1.6 you found either x <= -1.275 (roughly) so -1.6 <= x < -1.275 or x > 6.275 now you have to check the other condition, where x <= -8/5
mathmath333
  • mathmath333
for \(x<-8.5\) \(-\infty
phi
  • phi
in other word, when (5x-8) is negative, we can drop the absolute value sign, and negate the value inside. (for example, if we know we have a negative number, say -2 inside the abs value: | -2 | we know this is the same as - (-2) = +2 ) so x^2 - -(5x+8)>0 or x^2 +5x+8> 0 complete the square x^2 + 5x +25/4 +8 > 25/4 (x+5/2)^2 > 25/4 - 8 or \[ (x+5/2)^2 > -7/4 \] which will be true for all x (the left side will be 0 or bigger for any x) but we assumed x <= -8/5 so the solution is x <= -8/5
phi
  • phi
now combine -1.6 <= x < -1.275 or x > 6.275 and x<= -8/5 = -1.6 we get \[ -\infty < x < \frac{5-\sqrt{57}}{2} \text{ or } x >\frac{5+\sqrt{57}}{2} \]
phi
  • phi
Does this stuff make sense? If you are missing some idea, you should ask.
mathmath333
  • mathmath333
no question , thnx
mathmath333
  • mathmath333
how do u approximate \(\sqrt{57}\) without calculater
phi
  • phi
I don't. but I know its bigger than 7 (7*7= 49) and less than 8 (8*8=64)
phi
  • phi
I only used decimals because it was easier to type (and easier to compare to -8/5)
mathmath333
  • mathmath333
ok

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