## mathmath333 one year ago solve for $$x$$

1. mathmath333

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

2. misty1212

HI!!

3. misty1212

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

4. mathmath333

i worked m onfised

5. misty1212

??

6. mathmath333

*confused

7. misty1212

if $$x>-\frac{8}{5}$$ then $|5x+8|=5x+8$ right ?

8. misty1212

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

9. mathmath333

i dk how u got it

10. misty1212

the absolute value is a piece wise function

11. misty1212

let me write it as one hold on

12. mathmath333

ok i got it

13. 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 $x^2-5x-8>0$

14. 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 $x^2+5x+8>0$

15. misty1212

ach

16. mathmath333

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

17. misty1212

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

18. misty1212

looks good to me

19. mathmath333

lol m still confused

20. 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}}

21. mathmath333

@phi

22. mathmath333

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

23. 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

24. mathmath333

for $$x<-8.5$$ $$-\infty<x<-8/5$$

25. 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

26. phi

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

27. phi

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

28. mathmath333

no question , thnx

29. anonymous
30. mathmath333

how do u approximate $$\sqrt{57}$$ without calculater

31. phi

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

32. phi

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

33. mathmath333

ok