document.write( "Question 119712: Find three consecutive integers such that the sum of their squares is 77. \n" ); document.write( "

Algebra.Com's Answer #87712 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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x^2 + (x+1)^2 + (x+2)^2 = 77
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\n" ); document.write( "FOIL
\n" ); document.write( "x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) = 77
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\n" ); document.write( "Group like terms on the left:
\n" ); document.write( "x^2 + x^2 + x^2 + 2x + 4x + 1 + 4 - 77 = 0
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\n" ); document.write( "3x^2 + 6x - 72 = 0
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\n" ); document.write( "Simplify, divide by 3
\n" ); document.write( "x^2 + 2x - 24 = 0
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\n" ); document.write( "Factors to:
\n" ); document.write( "(x+6)(x-4) = 0
\n" ); document.write( "x = +4; then: 4, 5, 6 are the 3 consecutive digits
\n" ); document.write( "and
\n" ); document.write( "x = -6; then -6, -5, -4 are the 3 consecutive digits \n" ); document.write( "

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