dddan
how do sove for the solutions for ;
the square root of x+ the square root of x36 which then =0?



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tejasree
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Subtract the square root of x36 from both sides!
now: the square root of x = the square root of x36

tejasree
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\[\sqrt{x} = \sqrt{x36}\]

dddan
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what about the 2?

tejasree
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Square both of them\[(\sqrt{x})^{2} = (\sqrt{x36})^{2}\]

tejasree
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what 2

tejasree
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?

dddan
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im so sry the whole thing is equal to 2 not 0 im so sry

tejasree
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ok...then it is:
\[(\sqrt{x})^{2} = (2\sqrt{x36})^{2}\]

tejasree
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\[x = 4  2\sqrt{x36}  2\sqrt{x36} + x  36\]

tejasree
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\[x = 32 + x + 4\sqrt{x36}\]

tejasree
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\[x  x + 32 = 4\sqrt{x36}\]

tejasree
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\[32 = 4\sqrt{x36}\]

tejasree
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\[(32)^{2} = (4\sqrt{x36})^{2}\]

tejasree
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no no!!

tejasree
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wait a sec!

tejasree
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Instead of square rooting 32 and the other, first divide 4 from both sides!

tejasree
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\[(8)^{2} = (\sqrt{x36})^{2}\]

tejasree
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you following?

tejasree
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64 = X36
64 + 36 = X  36 + 36
100 = X

tejasree
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thank you! :)