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sqrt (x-2) + sqrt (x+2)=4 i have no idea how to get rid of the surd signs
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gotta square twice
first time get \[x-2+2\sqrt{(x-2)(x+2)}+x+2=16\]
giving \[2\sqrt{(x-2)(x+2)}=16-2x\] or \[\sqrt{(x+2)(x-2)}=8-x\] then square again
\[(x+2)(x-2)=64-16x+x^2\]
\[x^2-4=16-16x+x^2\] \[-4=16-16x\] \[16x=20\] \[x=\frac{5}{4}\] hmm did i mess this up? lets check
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oh yes i should be \[x^2-4=64-16x+x^2\]
i turned the 64 into a 16 my mistake
ok better. we get \[68=16x\] \[x=\frac{68}{16}=\frac{17}{4}\]
and check is \[\sqrt{\frac{17}{4}-2}+\sqrt{\frac{17}{4}+2}\] \[=\sqrt{\frac{9}{4}}+\sqrt{\frac{25}{4}}\] \[=\frac{3}{2}+\frac{5}{2}=\frac{8}{2}=4\]
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