## anonymous one year ago ques

1. anonymous

why is sqrt{x^2}=|x|?? let f(x)=sqrt{x^2} g(x)=x h(x)=|x| at x= -1 we have f(x)=plus minus 1 g(x)=-1 h(x)=1 clearly it looks like sqrt{x^2}=plus minus x sorry my equation box is giving errors right now so im using this

2. ganeshie8

|dw:1435127988303:dw|

3. anonymous

plus minus 2?

4. ganeshie8

Also |dw:1435128205480:dw|

5. anonymous

|dw:1435128362002:dw|

6. ganeshie8

we could say the "2nd roots of 4" are "+2 and -2" as both of them when rised to 2nd power give us 4; but if we stick to the definition of sqrt function, we only pick the positive root.

7. ikram002p

@Nishant_Garg what is x^2 range ?

8. ganeshie8

$$\sqrt{4} = 2$$

9. anonymous

I'm confused, doesn't square rooting gives 2 values?

10. ganeshie8

By definition, sqrt(x) is always the positive root of x

11. ganeshie8

sqrt(x) is a single valued function

12. anonymous

I don't understand, that sounds really really wrong

13. ganeshie8

do you have definition of sqrt(x) function wid you ?

14. ganeshie8

stick to that

15. anonymous

Well, idk, I've been doing plus minus x all the time for sqrt of x^2 How do you suppose the equation x^2=4 has 2 solutions then if u only take the positive value?

16. ganeshie8

$x^2=4 \implies x = \pm \sqrt{~4~} = \pm 2$

17. ganeshie8

$x^2 =3 \implies x=\pm\sqrt{3}$

18. ganeshie8

i think you're confusing sqrt(x) function with the complex roots of x

19. anonymous

This is melting my brain, it feels like I've been square rooting wrong the whole time... so is x^2=4 different from x=sqrt{4} ??

20. ganeshie8

Yes they are not same. $x = \sqrt{4} \implies x^2=4$ but the converse is not true.

21. ganeshie8

$x^2=4 \implies x=\color{red}{\pm}\sqrt{4}$

22. anonymous

and x=sqrt{4} implies x=2 ONLY ?

23. ganeshie8

there wont be any confusion if we take the definition seriosuly and stick to it

24. ganeshie8

that is correct $x=\sqrt{4}\implies x=2$

25. anonymous

and not minus 2

26. ganeshie8

Yep. sqrt(x) is never negative

27. anonymous

This is just bizzare, I always thought sqrt gives plus minus.....It all looks like a different point of view now

28. anonymous

then the equation x^2=4 is to be solved like sqrt{x^2}=sqrt{4} implies |x|=2(by definition of sqrt) removing the mod, x=plus minus 2 but for the equation x=sqrt{4} we have x=2(by definition of sqrt)

29. ganeshie8

Haha ikr! one not so cool thing about mathematics is sometimes we just need to follow rules and stick to definitions even though they seem unintuitive

30. anonymous

can u check my post

31. ganeshie8

that looks good to me!

32. anonymous

so the plus minus is a consequence of the modulus function and not square root

33. ganeshie8

modulus function is a consequence of sqrt

34. ganeshie8

|dw:1435129510602:dw|

35. ganeshie8

as you said |x|=2 has two solutions

36. anonymous

makes sense now but feels a little different :)

37. ganeshie8

since we're messing with roots, notice that below is perfectly okay : |dw:1435129597807:dw|

38. ganeshie8

but below is wrong : |dw:1435129708335:dw|

39. anonymous

should be |X^7|?

40. ganeshie8

Yes! |x^7| or |x|^7 both are fine

41. anonymous

as it has an odd power it can gives negative numbers

42. ganeshie8

Exactly! which changes the original function sqrt(x^14) as x^14 is always positive

43. anonymous

amazing thanks a lot...

44. ganeshie8

np:)