document.write( "Question 753630: find x, if;
\n" ); document.write( "\"+%285%2B2sqrt2%29%5E%28x%5E2-x%29+%2B+%285-2sqrt2%29%5E%28x%5E2-x%29+=10++\"
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Algebra.Com's Answer #458572 by KMST(5328)\"\" \"About 
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Consider the function \"g%28y%29=%285%2B2sqrt%282%29%29%5Ey%2B%285-2sqrt%282%29%29%5Ey\"
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\n" ); document.write( "\"g%28y%29=%285%2B2sqrt2%29%5Ey%2B%285-2sqrt2%29%5Ey\" is the sum of two exponential functions,
\n" ); document.write( "with positive bases, \"0%3C5-2sqrt2%3C5%2B2sqrt2\".
\n" ); document.write( "Those two exponmential functions, and \"g%28y%29\", increase with \"y\" throughout their real numbers domain, from -infinity to infinity.
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\n" ); document.write( "So for \"y%3C1\" --> \"g%28y%29%3C10\"
\n" ); document.write( "and for \"y%3E1\" --> \"g%28y%29%3E10\"
\n" ); document.write( "The only value of \"y\" that makes \"g%28y%29=10\" is \"y=1\"
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\n" ); document.write( "Then, the only solutions for \"%285%2B2sqrt%282%29%29%5E%28x%5E2-x%29%2B%285-2sqrt%282%29%29%5E%28x%5E2-x%29=10\"
\n" ); document.write( "will be the solutions of \"x%5E2-x=1\"
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\n" ); document.write( "\"x%5E2-x=1\" --> \"x%5E2-x-1=0\"
\n" ); document.write( "Applying the quadratic formula, we find
\n" ); document.write( "\"x=%28-1+%2B-+sqrt%281%5E2-4%281%29%28-1%29%29%29%2F2%2F1\" --> \"highlight%28x=%281+%2B-+sqrt%285%29%29%2F2%29\"\r
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