document.write( "Question 85836: the square root of 2x plus three minus the square root ofx plus one equals one \n" ); document.write( "
Algebra.Com's Answer #62052 by ankor@dixie-net.com(22740)\"\" \"About 
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the square root of 2x plus three minus the square root ofx plus one equals one
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\n" ); document.write( "Here it is step-by-step, the idea is to perform algebra operations that will get rid of the radicals; Leaving us with a quadratic equation that we solve
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\n" ); document.write( "\"sqrt%282x+%2B+3%29\" - \"sqrt%28x%2B1%29\" = 1
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\n" ); document.write( "add \"sqrt%28x%2B1%29\" to both sides
\n" ); document.write( "\"sqrt%282x+%2B+3%29\" = \"sqrt%28x%2B1%29\" + 1
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\n" ); document.write( "Square both sides:
\n" ); document.write( "2x + 3 = (x+1) + 2*sqrt(x+1) + 1; FOILed [(sqrt(x+1) + 1] * [sqrt(x+1) + 1]
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\n" ); document.write( "2x + 3 = x + 2 + 2*sqrt(x+1)
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\n" ); document.write( "2x - x + 3 - 2 = 2*sqrt(x+1)
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\n" ); document.write( "x + 1 = 2*Sqrt(x+1)
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\n" ); document.write( "Square both sides again:
\n" ); document.write( "x^2 + 2x + 1 = 4(x+1); FOILed (x+1)*(x+1)
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\n" ); document.write( "x^2 + 2x + 1 = 4x + 4
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\n" ); document.write( "x^2 + 2x - 4x + 1 - 4 = 0
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\n" ); document.write( "x^2 - 2x - 3 = 0
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\n" ); document.write( "Factors to:
\n" ); document.write( "(x - 3)(x + 1 ) = 0
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\n" ); document.write( "x = +3
\n" ); document.write( "and
\n" ); document.write( "x = -1
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\n" ); document.write( "Check both solutions by substitution in the original equation.
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