document.write( "Question 750840: hi please help me with this question: i think it's under quadratic equation im not sure.\r
\n" ); document.write( "\n" ); document.write( "Show that the real solution of the equation ax2+bx+c=0 are the reciprocals of the real solutions of the equation cx2+bx+a=0. Assume that b2-4ac≥0\r
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Algebra.Com's Answer #456840 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
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document.write( " ax²+bx+c=0\r\n" );
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document.write( "The solutions are\r\n" );
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document.write( "\"%28-b+%2B+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\" and \"%28-b+-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Ac%29+\" \r\n" );
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document.write( " cy²+by+a=0\r\n" );
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document.write( "The solutions are\r\n" );
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document.write( "\"%28-b+%2B+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Ac%29+\" and \"%28-b+-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Ac%29+\"\r\n" );
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document.write( "They are reciprocals then their product must be 1:\r\n" );
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document.write( "Let's multiply the solution of the first with a positive radical times\r\n" );
document.write( "the solution of the second with a negative radical:\r\n" );
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document.write( "\"%28%28-b+%2B+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29%29+\"\"%22%22%2A%22%22\"\"%28%28-b+-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Ac%29+%29%29\" \r\n" );
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document.write( "The two factors in the numerators are conjugates so FOILing them\r\n" );
document.write( "causes outers and inners to cancel, so we get:\r\n" );
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document.write( "\"%28%28-b%29%5E2+-+%28sqrt%28b%5E2-4ac%29%29%5E2%29%2F%284ac%29+\" =\r\n" );
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document.write( "\"%28b%5E2-%28b%5E2-4ac%29%29%2F%284ac%29\" =\r\n" );
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document.write( "\"%28b%5E2-b%5E2%2B4ac%29%2F%284ac%29\" =\r\n" );
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document.write( "\"%284ac%29%2F%284ac%29\" =\r\n" );
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document.write( "1\r\n" );
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document.write( "so they are reciprocals.  Multiplying the other pair\r\n" );
document.write( "gives the same results.\r\n" );
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document.write( "Edwin
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