document.write( "Question 818245: In one root ax^2+bX+c=0 is double the other.,prove that 2b^2=9ac ?????? \n" ); document.write( "

Algebra.Com's Answer #492444 by KMST(5328) $\"\"$ $\"About$

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The solutions to $\"ax%5E2%2Bbx%2Bc=0+\"$ are $\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"$ ,
\n" ); document.write( "which I will temporarily abbreviate as $\"x+=+%28-b+%2B-+sqrt%28DELTA%29%29%2F%282%2Aa%29+\"$
\n" ); document.write( "with $\"DELTA=b%5E2-4%2Aa%2Ac\"$ .
\n" ); document.write( "
\n" ); document.write( "If $\"%28-b%2Bsqrt%28DELTA%29%29%2F%282%2Aa%29=+2%28%28-b-sqrt%28DELTA%29%29%2F%282%2Aa%29%29\"$ ,
\n" ); document.write( " $\"-b%2Bsqrt%28DELTA%29=+2%28-b-sqrt%28DELTA%29%29\"$
\n" ); document.write( " $\"-b%2Bsqrt%28DELTA%29=+-2b-2sqrt%28DELTA%29\"$
\n" ); document.write( " $\"3sqrt%28DELTA%29=+-b\"$ which squares into $\"9DELTA=b%5E2\"$ .
\n" ); document.write( "
\n" ); document.write( "If $\"%28-b-sqrt%28DELTA%29%29%2F%282%2Aa%29=+2%28%28-b%2Bsqrt%28DELTA%29%29%2F%282%2Aa%29%29\"$ ,
\n" ); document.write( " $\"-b-sqrt%28DELTA%29=+2%28-b%2Bsqrt%28DELTA%29%29\"$
\n" ); document.write( " $\"-b-sqrt%28DELTA%29=+-2b%2B2sqrt%28DELTA%29\"$
\n" ); document.write( " $\"-b=3sqrt%28DELTA%29\"$ which squares into $\"b%5E2=9DELTA\"$ .
\n" ); document.write( "
\n" ); document.write( "Either way $\"b%5E2=9DELTA\"$ which means $\"b%5E2=9%28b%5E2-4%2Aa%2Ac%29\"$ .
\n" ); document.write( "So $\"b%5E2=9b%5E2-36%2Aa%2Ac%29\"$ , and adding $\"36%2Aa%2Ac-b%5E2\"$ to both sides we get
\n" ); document.write( " $\"36%2Aa%2Ac=8b%5E2%29\"$ , which dividing both sides by 4 gives us
\n" ); document.write( " $\"9ac=2b%5E2%29\"$ or $\"highlight%282b%5E2=9ac%29\"$ . \n" ); document.write( "

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