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anonymous

  • 5 years ago

(√x+10 ) + (√x-6) = 8 pls show work

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  1. anonymous
    • 5 years ago
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    Add like terms

  2. anonymous
    • 5 years ago
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    2 sqrt[x]+10-6

  3. anonymous
    • 5 years ago
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    2 sqrt[x]+4=8 2 sqrt[x]=4 (sqrt[x])^2=(2)^2 x=4

  4. Yuki
    • 5 years ago
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    littleice, is this the problem you need help with ?

  5. anonymous
    • 5 years ago
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    i know the answer comes to 15

  6. anonymous
    • 5 years ago
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    yes thanks

  7. Yuki
    • 5 years ago
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    let me ask you something first is it \[\sqrt x +10\] or \[\sqrt{x+10}\]

  8. anonymous
    • 5 years ago
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    \[\sqrt{x+10}+\sqrt{x-6}=8\]

  9. Yuki
    • 5 years ago
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    okay, so our main goal is to get rid of the square roots. but before we proceed, there are things that you want to always remember about eqn.s that involves sqrt. it is that there are x's that are not allowed to use. since te number inside the radical has to be positive, first we have to note that x>-10 and x>6

  10. anonymous
    • 5 years ago
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    i thought it was x cant equal -10 or 6?

  11. Yuki
    • 5 years ago
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    if we get an answer that doesn't fall in this category, then there are no solutions. another thing that you have to do, (I'm not going to go into the detail unless you want to) is to check the answers when you get it. the gist of the reason has to do with the fact that\[x^2=k, means, x= \pm \sqrt{k}\]

  12. Yuki
    • 5 years ago
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    -10 and 6 are actually ok. because sqrt(0) = 0 but sqrt(x) when x<0, becomes and imaginary number

  13. anonymous
    • 5 years ago
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    k

  14. Yuki
    • 5 years ago
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    alright let's proceed

  15. Yuki
    • 5 years ago
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    when you get rid of sqrt.s you want to square them. but when you square both sides just as it is, you will end up having a square root again, so to avoid that, you will isolate one of the terms that has a square root around it, and deal with that first. as follows

  16. Yuki
    • 5 years ago
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    \[(\sqrt{x+10})^2 = (8-\sqrt{x-6})^2\] so\[x+10 = 64 +16\sqrt{x-6}+(x-6)\]

  17. Yuki
    • 5 years ago
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    now you only have one square root, so you isolate it and then square both sides again \[({x + 10 -64 -x +6 \over 16})^2 = \sqrt{x-6}^2\]

  18. Yuki
    • 5 years ago
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    that way you will have a quadratic eqn. and you know how to solve it from there. don't forget to check the solutions, by plugging them in because sometimes they will fail.

  19. Yuki
    • 5 years ago
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    \[(-3 )^2= \sqrt(x-6)^2\] is what you will get when you simplify the above so 9 = x -6 thus x =15

  20. Yuki
    • 5 years ago
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    however, let's see what happens when we plug this back in into the original eqn. sqrt(15+10)+sqrt(15-6) =5+3 =8 which actually works. now we can guarantee that x =15.

  21. Yuki
    • 5 years ago
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    did it help at all ?

  22. anonymous
    • 5 years ago
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    idk what went thru my head but i got confused on how u did it did it a weird way and got the answer aswell

  23. Yuki
    • 5 years ago
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    oh... ok :( let me know if you need any explanations though :)

  24. anonymous
    • 5 years ago
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    heres how i did it \[\left( \sqrt{x+10} \right)^{2} = (8-\sqrt{x-6})^{2}\] \[x+10=64+16\sqrt{x-6} + \sqrt{x-6}\] \[x+10-64-x-6=16\sqrt{x-6}\] \[48^{2}=(16\sqrt{x-6})^{2}\] 2304=256x-6 2304/256=x-6 9=x-6 15=x

  25. Yuki
    • 5 years ago
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    it is the same thing

  26. Yuki
    • 5 years ago
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    I just divided 16 on both sides first

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