document.write( "Question 282842: if p and q are the zeroes of a polynomial ax2+bx+c then evaluate p-q \n" ); document.write( "
Algebra.Com's Answer #205405 by Edwin McCravy(20065)\"\" \"About 
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document.write( "\"ax%5E2%2Bbx%2Bc\" where \"a%3C%3E0\"\r\n" );
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document.write( "The sum of the zeros is \"p%2Bq=-b%2Fa\"\r\n" );
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document.write( "The product of the zeros is \"pq=c%2Fa\"\r\n" );
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document.write( "We will assume the maximal zero is p and the minimal zero is q,\r\n" );
document.write( "so that p-q will be non-negative, though it may be zero.\r\n" );
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document.write( "We notice that the square of the sum of the zeros has similar terms\r\n" );
document.write( "to the square of the difference. Notice that:\r\n" );
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document.write( "\"%28p%2Bq%29%5E2=p%5E2%2B2pq%2Bq%5E2\" is very much like \"%28p-q%29%5E2=p%5E2-2pq%2Bq%5E2\"\r\n" );
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document.write( "except that the first has a \"%22%22%2B2pq\" term whereas the second has \r\n" );
document.write( "a \"-2pq\" term.  \r\n" );
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document.write( "To get an expression for \"p-q\" we start with this:\r\n" );
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document.write( "\"p-q=sqrt%28%28p-q%29%5E2%29=sqrt%28p%5E2-2pq%2Bq%5E2%29\"\r\n" );
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document.write( "Let's create the square of the sum of the zeros under the\r\n" );
document.write( "radical by adding and then subtracting the term \"red%282pq%29\",\r\n" );
document.write( "which does not change the value since this amounts to adding 0.\r\n" );
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document.write( "\"p-q=sqrt%28%28p-q%29%5E2%29=sqrt%28p%5E2-2pq%2Bq%5E2%2Bred%282pq%29-red%282pq%29%29\"\r\n" );
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document.write( "Swapping two of the terms under the radical:\r\n" );
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document.write( "\"p-q=sqrt%28%28p-q%29%5E2%29=sqrt%28p%5E2%2Bred%282pq%29%2Bq%5E2-2pq-red%282pq%29%29\"\r\n" );
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document.write( "Factoring the first three terms under the radical and combining the\r\n" );
document.write( "last two terms:\r\n" );
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document.write( "\"p-q=sqrt%28%28p-q%29%5E2%29=sqrt%28%28p%2Bq%29%5E2-4pq%29\"\r\n" );
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document.write( "Now since \"p%2Bq=-b%2Fa\" we substitute \"-b%2Fa\" for \"p%2Bq\" and\r\n" );
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document.write( "and since \"pq=c%2Fa\" we substitute \"c%2Fa\" for \"pq\":\r\n" );
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document.write( "\"p-q=sqrt%28+%28-b%2Fa%29%5E2-4%28c%2Fa%29%29\"\r\n" );
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document.write( "\"p-q=sqrt%28%28b%5E2%29%2F%28a%5E2%29-%284c%29%2Fa%29\"\r\n" );
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document.write( "We get a common denominator of \"a%5E2\" under the radical, so\r\n" );
document.write( "we multiply the second term under the radical by \"a%2Fa\"\r\n" );
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document.write( "\"p-q=sqrt%28%28b%5E2%29%2F%28a%5E2%29-%284c%2Aa%29%2F%28a%2Aa%29%29\"\r\n" );
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document.write( "\"p-q=sqrt%28%28b%5E2%29%2F%28a%5E2%29-%284ac%29%2F%28a%5E2%29%29\"\r\n" );
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document.write( "Combine the fractions over the LCD:\r\n" );
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document.write( "\"p-q=sqrt%28%28b%5E2-4ac%29%2F%28a%5E2%29%29\"\r\n" );
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document.write( "Taking square roots of numerator and denominator:\r\n" );
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document.write( "\"p-q=sqrt%28b%5E2-4ac%29%2Fsqrt%28a%5E2%29\"\r\n" );
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document.write( "Since \"a\" may be negative, we must use absolute value of a,\r\n" );
document.write( "since p-q is non-negative:\r\n" );
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document.write( "\"p-q=sqrt%28b%5E2-4ac%29%2Fabs%28a%29\"\r\n" );
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document.write( "Edwin
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