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## anonymous one year ago I need help with algebra again

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1. anonymous

$-\sqrt{6x+1}-5=-7-\sqrt{3x+1}$

2. UsukiDoll

thinking

3. anonymous

lol ok

4. UsukiDoll

ok get the numbers to the right and those square roots to the left

5. Michele_Laino

hint: we can rewrite your equation as below: $\Large 7 - 5 = \sqrt {6x + 1} - \sqrt {3x + 1}$

6. UsukiDoll

half of me is like square both sides to get rid of those square roots. after that step

7. Michele_Laino

after a first square, we get: $\Large 4 = 6x + 1 + 3x + 1 - 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)}$ please simplify @MathHater82

8. UsukiDoll

oh golly .

9. anonymous

o.o im lost

10. Michele_Laino

hint: here are the next steps: $\Large \begin{gathered} 4 = 9x + 2 - 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} \hfill \\ \hfill \\ 2 = 9x - 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} \hfill \\ \end{gathered}$

11. anonymous

ok but how did you do the step before that

12. UsukiDoll

$\Large 2= \sqrt {6x + 1} - \sqrt {3x + 1}$ $\Large 2^2= (\sqrt {6x + 1} - \sqrt {3x + 1})^2$ $\Large 4= (\sqrt {6x + 1} - \sqrt {3x + 1})(\sqrt {6x + 1} - \sqrt {3x + 1})$ and then expand ?

13. UsukiDoll

with foil?

14. Michele_Laino

since I have computed the square of both sides, like this: $\Large {\left( {7 - 5} \right)^2} = {\left( {\sqrt {6x + 1} - \sqrt {3x + 1} } \right)^2}$

15. anonymous

oh ok

16. UsukiDoll

good luck to that on the right hand side that's killer

17. Michele_Laino

next, we can write: $\Large 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} = 9x - 2$

18. Michele_Laino

and, taking the square of both sides again, we have: $\Large {\left( {2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} } \right)^2} = {\left( {9x - 2} \right)^2}$

19. Michele_Laino

or, after a simplification: $\Large 4\left( {18{x^2} + 9x + 1} \right) = 81{x^2} + 4 - 36x$

20. Michele_Laino

I have applied the foil method inside the radical at left side

21. Michele_Laino

so, please continue @MathHater82

22. UsukiDoll

oh yes it's getting to be a bit easier. . . . after those square roots are gone.

23. anonymous

ok im just trying to follow this all on my white board i have here

24. anonymous

give me a sec

25. anonymous

$36x+4=9x^2-36x+4$ $72x=9x^2$

26. Michele_Laino

ok! correct! now try to solve that equation

27. anonymous

x=8

28. Michele_Laino

hint: we have the subsequent steps: $\Large \begin{gathered} 9{x^2} - 72x = 0 \hfill \\ \hfill \\ 9x\left( {x - 8} \right) = 0 \hfill \\ \end{gathered}$

29. UsukiDoll

maybe we can shift the 9x^2 and se the equation equal to 0. whoops backwards 0 = 9x^2-72x

30. UsukiDoll

yes solve for x after what @Michele_Laino typed there's 2 x's to solve.. so first we have 9x=0 and then x-8=0

31. Michele_Laino

x=8 is correct! nevertheless also x=0 is a solution that's right! @UsukiDoll

32. UsukiDoll

x = 0,8 but uh oh only one x works

33. Michele_Laino

so, x=0 is a solution or not? @MathHater82

34. Michele_Laino

please recall this step: $\Large 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} = 9x - 2$

35. UsukiDoll

I think plugging in x = 0 in the equation that we had started with is easier.

36. Michele_Laino

noew, the left side is positive, since it is the positive square root, so also the right side has to be positive. Namely we have: $\Large 9x - 2 > 0$

37. Michele_Laino

now*

38. Michele_Laino

or: $\Large x > \frac{9}{2}$ now x=0, doesn't check that condition, so x=0 can not be an acceptable solution

39. Michele_Laino

|dw:1437567359392:dw|

40. UsukiDoll

also when x = 0 $-\sqrt{1}-5=-7-\sqrt{1}$ -1-5=-7-1 $-6 \neq -8$

41. Michele_Laino

yes! that is another possible way to solve the question

42. Michele_Laino

mathematicians call x=0 an extraneous solution

43. anonymous

I don't really understand why x=0

44. Michele_Laino

since x=0 and x=8 are the solutions of this equation: $\Large 9{x^2} - 72x = 0$

45. UsukiDoll

we obtained two solutions, but 1 works.

46. anonymous

how did we obtain the solution 0 and which one worked

47. UsukiDoll

the equation $\Large 9{x^2} - 72x = 0$ got factored.. we can factor a 9x out

48. UsukiDoll

$\large 9x(x-8)=0$ then we solve for each case 9x=0 and x-8=0 that leads us to x =0,8 but only 1 works.

49. anonymous

ok

50. Michele_Laino

furthermore. looking at your original equation also these conditions have to be checked: $\Large \begin{gathered} 6x + 1 \geqslant 0 \hfill \\ 3x + 1 \geqslant 0 \hfill \\ \end{gathered}$ for existence of both radicals, which are automatically satisfied since: $\Large 8 > \frac{9}{2}$

51. anonymous

ok cool i get it thanks for the help

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