Question

solve for \(x\)

Answer 1

\(\large \color{black}{\begin{align} x^2-|5x+8|>0\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

HI!!

Answer 3

work in cases
solve if \(x<-\frac{8}{5}\) and if \(x>-\frac{8}{5}\)

Answer 4

i worked m onfised

Answer 5

??

Answer 6

*confused

Answer 7

if \(x>-\frac{8}{5}\) then
\[|5x+8|=5x+8\] right ?

Answer 8

if it is not clear how i got that say so and i will explain

Answer 9

i dk how u got it

Answer 10

the absolute value is a piece wise function

Answer 11

let me write it as one hold on

Answer 12

ok i got it

Answer 13

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
\[x^2-5x-8>0\]

Answer 14

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 \[x^2+5x+8>0\]

Answer 15

ach

Answer 16

\(\large \color{black}{\begin{align} x>\dfrac{5+\sqrt{57}}{2},\ x<\dfrac{5-\sqrt{57}}{2} \hspace{.33em}\\~\\
\end{align}}\)

Answer 17

intersection of your solution and \(x>-\frac{8}{5}\)

Answer 18

looks good to me

Answer 19

lol m still confused

Answer 20

this is solution of

\(x^2-5x-8>0\)

\(\large \color{black}{\begin{align} x>\dfrac{5+\sqrt{57}}{2},\ x<\dfrac{5-\sqrt{57}}{2} \hspace{.33em}\\~\\
\end{align}}\)

Answer 21

@phi

Answer 22

for \(x^2+5x+8>0\)
\(\large \color{black}{\begin{align} -\infty<x<+\infty \hspace{.33em}\\~\\
\end{align}}\)

Answer 23

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

Answer 24

for \(x<-8.5\)

\(-\infty<x<-8/5\)

Answer 25

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

Answer 26

now combine
-1.6 <= x < -1.275 or x > 6.275
and x<= -8/5 = -1.6
we get
\[ -\infty < x < \frac{5-\sqrt{57}}{2} \text{ or } x >\frac{5+\sqrt{57}}{2} \]

Answer 27

Does this stuff make sense? If you are missing some idea, you should ask.

Answer 28

no question , thnx

Answer 29

http://openstudy.com/study#/updates/5593fc3ae4b0989cc8780f4d

Answer 30

how do u approximate \(\sqrt{57}\) without calculater

Answer 31

I don't. but I know its bigger than 7 (7*7= 49) and less than 8 (8*8=64)

Answer 32

I only used decimals because it was easier to type (and easier to compare to -8/5)

Answer 33

ok