document.write( "Question 1161183: Please help me solve this:
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\n" ); document.write( "The quadratic equation {{ (b-c)x^2+(c-a)x+(a-b)=0 }}} has a repeated real solution. Prove that \"+b=%28a%2Bc%29%2F2+\"
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Algebra.Com's Answer #784670 by math_helper(2461)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Consider  \"Ax%5E2+%2B+Bx+%2B+C+=+0+\"  (using capital letters to distinguish the coefficients from the given problem).\r
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document.write( "If one applies the quadratic formula, they get the roots:\r
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document.write( "     \"+x+=+%28-B+%2B-+sqrt%28B%5E2+-+4AC%29%29%2F%282A%29+\"\r
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document.write( "The roots are real and repeated when  \"+sqrt%28B%5E2-4AC%29+=+0+\" or we can say when \"+B%5E2+-+4AC+=+0\"   (this is called the discriminant).\r
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document.write( "Compare A,B,C to the given problem:\r
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document.write( "  A = b-c
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document.write( "  B = c-a
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document.write( "  C = a-b\r
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document.write( "Now use the right hand sides of these to get the discriminant:\r
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document.write( "       \"+%28c-a%29%5E2+-+4%28b-c%29%28a-b%29+=+0+\"\r
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document.write( "This can be expanded, then simplified:
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document.write( "       \"+c%5E2-2ac%2Ba%5E2++-4ab%2B4b%5E2%2B4ac-4bc+=+0+\"
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document.write( "       \"+c%5E2%2B2ac%2Ba%5E2++%2B4b%5E2+-+4ab-4bc++=+0+\"
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document.write( "       \"+4b%5E2+-+4%28a%2Bc%29b+%2B+%28a%2Bc%29%5E2+=+0+\"\r
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document.write( "divide thru by 4:
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document.write( "       \"+b%5E2+-+%28a%2Bc%29b+%2B+%28a%2Bc%29%5E2%2F4+=+0+\"\r
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document.write( "       \"+b%5E2+-+%28a%2Bc%29b+%2B+%28%28a%2Bc%29%2F2%29%5E2+=++0+\"\r
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document.write( "Notice this is a perfect square:
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document.write( "        \"+%28b+-+%28a%2Bc%29%2F2%29%5E2+=+0+\"\r
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document.write( "Which has the solution:
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document.write( "        \"+b+=+%28a%2Bc%29%2F2+\"\r
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document.write( "Recall:
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document.write( "It is this relationship between a,b,and c that must hold in order for the original descriminant to be zero.   That in turn, implies repeated real roots.  
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