## amorfide 3 years ago Soooooo I need help with a modulus/absolute value equation

1. amorfide

|dw:1350320124515:dw|

2. amorfide

any method just explain

3. amorfide

@ParthKohli

4. amorfide

|dw:1350320261447:dw| my graph

5. zugzwang

If |x| = y, then x = y OR x = -y

6. amorfide

yes i get that part, so this gives |dw:1350321371149:dw|

7. amorfide

then figure out x in both cases but i will get 4 answers, it only crosses 3 times, how do i know which one to eliminate?

8. 2le

|dw:1350321424876:dw| what you drew looks more exponential. The graph for y=|4/x| is linear with a jump at x=0.

9. zugzwang

There must be one for which x/2 + 3 is negative. That's an impossibility.

10. zugzwang

Because x/2 + 3 is the absolute value of something...?

11. amorfide

|dw:1350321988874:dw| |dw:1350322071606:dw|

12. amorfide

so what would my final answers be

13. zugzwang

Are you sure about the first part?

14. zugzwang

I think it should be -3 (+ or -) sqrt (21)

15. zugzwang

16. zugzwang

in any case, -3-sqrt(17), try substituting that for x in x/2 + 3

17. zugzwang

And remember -3-sqrt(17) < -3-sqrt(16) = -7

18. amorfide

thank you!

19. zugzwang

No problem... Sorry about my pathetic arithmetic :P

20. amorfide

still made sense in a way :P

21. ParthKohli

$\rm \left|{4 \over x}\right| = {x \over 2} + 3 \qquad \Rightarrow \qquad {4 \over x} = \pm\left({x \over 2 }+3 \right)$Let's solve both.$\rm {4 \over x}={x \over 2}+3$so$8 = \rm x^2 + 6x$therefore$\rm x^2 + 6x-8=0$

22. ParthKohli

For the other solution,${4 \over \rm x } = \rm -{x \over 2}-3$so$8=-\rm x^2 - 6x$therefore$\rm x^2 +6x +8=0$

23. terenzreignz

And as @zugzwang was kind enough to point out, make sure all values of x lead to $\frac{x}{2}+3$ being nonnegative.

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