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anonymous
 5 years ago
(√x+10 ) + (√x6) = 8 pls show work
anonymous
 5 years ago
(√x+10 ) + (√x6) = 8 pls show work

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02 sqrt[x]+4=8 2 sqrt[x]=4 (sqrt[x])^2=(2)^2 x=4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0littleice, is this the problem you need help with ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i know the answer comes to 15

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me ask you something first is it \[\sqrt x +10\] or \[\sqrt{x+10}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{x+10}+\sqrt{x6}=8\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i thought it was x cant equal 10 or 6?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if 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\[x^2=k, means, x= \pm \sqrt{k}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.010 and 6 are actually ok. because sqrt(0) = 0 but sqrt(x) when x<0, becomes and imaginary number

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright let's proceed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(\sqrt{x+10})^2 = (8\sqrt{x6})^2\] so\[x+10 = 64 +16\sqrt{x6}+(x6)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now you only have one square root, so you isolate it and then square both sides again \[({x + 10 64 x +6 \over 16})^2 = \sqrt{x6}^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(3 )^2= \sqrt(x6)^2\] is what you will get when you simplify the above so 9 = x 6 thus x =15

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0however, let's see what happens when we plug this back in into the original eqn. sqrt(15+10)+sqrt(156) =5+3 =8 which actually works. now we can guarantee that x =15.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0idk what went thru my head but i got confused on how u did it did it a weird way and got the answer aswell

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh... ok :( let me know if you need any explanations though :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0heres how i did it \[\left( \sqrt{x+10} \right)^{2} = (8\sqrt{x6})^{2}\] \[x+10=64+16\sqrt{x6} + \sqrt{x6}\] \[x+1064x6=16\sqrt{x6}\] \[48^{2}=(16\sqrt{x6})^{2}\] 2304=256x6 2304/256=x6 9=x6 15=x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I just divided 16 on both sides first
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