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michelle1503
can anyone help me to the answer for this... (will post equation in a sec...)
\[\sqrt{\sqrt{x-2}+x}=2\]
root(x-2) + x =4 root(x-2) = 4-x square again
x-2 = 16 - 8x + x^2 9x = 18 + x^2 x^2 - 9x + 18 = 0
(x-6)(x-3)=0 x=6 or x=3
@niravshah08 how on earth did you get the first line x-2 = 16 - 8x + x^2 ???
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okay i understand.... but is it wrong if i do this.....
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and if it is wrong please explain why
but there cant be two different ways of doing it... i think yours is right... hmm
Michelle, your method is wrong, because: \[(a + b)^2 \neq a^2 + b^2\]
So, when squaring an equation, you literally put the whole LHS and RHS in huge brackets and square the whole bracket
so what your saying is that i can do this:
Also, at the end, always check your answers, because (usually when there's a square root) sometimes they yield a negative number under the square root, so the answer is not real
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yeah when i did it my method i got a random answer but did notice the - :) just my method is wrong LOL
Yeah that's correct. Although, it would be easier to, before you square it, take the x over to the other side. It just makes it easier to solve.
ah yeah i see i just tried it and it is possible but just involves alot of roots... and i dont like roots... they scare me... D': thanks for the explanation :D lifesaver :)
haha yeah, it gets a bit messy.. it's a pleasure :)