## pratu043 Find x. one year ago one year ago

1. pratu043

|dw:1342256143778:dw|

2. waterineyes

Try by rationalization process...

3. georgeblue22

|dw:1342256275514:dw|

4. waterineyes

Multiply and divide both the sides by: $\sqrt{x+4}-\sqrt{x -3}$

5. pratu043

You mean multiply by $(\sqrt{x+4} - \sqrt{x-3})/(\sqrt{x+4} - \sqrt{x-3})$

6. waterineyes

Yes...

7. ganeshie8

check @georgeblue22 once.. its very simple

8. pratu043

|dw:1342256415787:dw|

9. ganeshie8

i mean check @georgeblue22 reply once

10. waterineyes

No no don't do this...

11. pratu043

12. waterineyes

just do what @georgeblue22 said.. I was thinking it as an expression but it is an equation.. I forgot to see there 1 on right hand side...

13. TheViper

$\Huge{\color{gold}{\star \star}{\color{orange}{\text{Rationlise}}}}$

14. TheViper

best hint

15. pratu043

Oh yes, he brought the denominator to the other side.

16. TheViper

yes @pratu043

17. waterineyes

See, multiply both the sides by : $$\sqrt{x+4}+\sqrt{x -3}$$..

18. pratu043

That will help to get of the square root on the left side.

19. TheViper

Understood @pratu043 ??? :)

20. waterineyes

You will left with only one square root term.... after solving..

21. pratu043

Left side will be: |dw:1342256668499:dw|

22. waterineyes

No you are doing wrong..

23. pratu043

What's the mistake?

24. georgeblue22

|dw:1342256752422:dw|

Did you try squaring on both the sides ?

26. waterineyes

See show me what you get after multiply the term I said above... Do it slowly...

27. pratu043

Which term?

28. pratu043

So 7 = sqrt(x + 4) + sqrt(x - 3).

29. waterineyes

$\frac{(\sqrt{x+4} - \sqrt{x-3})}{(\sqrt{x+4} + \sqrt{x-3})} \times (\sqrt{x+4} + \sqrt{x-3}) = 1 \times (\sqrt{x+4} + \sqrt{x-3})$ Now solve this..

30. waterineyes

You will be left with: $\sqrt{x+4}- \sqrt{x-3} = \sqrt{x+4} + \sqrt{x-3}$ Now look it properly what is the term that is getting cancelling on both the sides???

31. pratu043

sqrt(x + 4).

32. waterineyes

Yes then cancel it and tell me what you are left with now..

33. pratu043

|dw:1342257083455:dw|

34. waterineyes

Can you bring them on one side now??

35. pratu043

|dw:1342257134878:dw|

36. waterineyes

37. pratu043

|dw:1342257180914:dw|

38. waterineyes

Yes... Have faith you are in a right direction.. Now squaring both the sides what will you get??

39. TheViper

sqaure both sides @pratu043

40. pratu043

|dw:1342257244043:dw|

41. waterineyes

And then put the brackets = 0 and find x from here..

42. pratu043

|dw:1342257276476:dw|

43. waterineyes

Yes.. You are right...

44. TheViper

yes @pratu043 u got it ;) deserve a medal ;)

45. pratu043

Thanks!!!

46. TheViper

yw ;)

47. waterineyes

See @pratu043 if we are given with: $\large (x-1)(y-2) = 0$ then you can do directly this: $(x - 1) =0 \quad and \quad (y - 2) = 0$ Getting???

48. waterineyes

So at the end we are left with: $4(x-3) = 0$ As $$4 \ne 0$$ So, You can put directly: $(x-3) = 0$ $x = 3$

49. pratu043

Oh I see .... I never knew that.

50. waterineyes

But this is possible in case we have 0 in right hand side.. Be careful...