document.write( "Question 1005804: one root of the equation 2x2-x+c=0 is double the other root. find c \n" ); document.write( "

Algebra.Com's Answer #621956 by MathLover1(20849) $\"\"$ $\"About$

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\n" ); document.write( "given:
\n" ); document.write( "one root of the equation $\"2x%5E2-x%2Bc=0\"$ is double the other root
\n" ); document.write( " find $\"c+\"$ \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "let one root be $\"alpha\"$
\n" ); document.write( "then the other root will be $\"2alpha\"$ \r
\n" ); document.write( "\n" ); document.write( "recall that sum of roots: $\"-b%2Fa\"$ \r
\n" ); document.write( "\n" ); document.write( "so, $\"alpha%2B2alpha=-b%2Fa\"$
\n" ); document.write( " $\"3alpha=+-b%2Fa\"$ ....since in $\"2x%5E2-x%2Bc=0\"$ , $\"a=2\"$ and $\"b=-1\"$ we have $\"3alpha=+-%28-1%29%2F2\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"3alpha=+1%2F2\"$
\n" ); document.write( " $\"alpha=+1%2F%282%2A3%29\"$
\n" ); document.write( " $\"alpha=+1%2F6\"$ .........one root
\n" ); document.write( " $\"2alpha=+2%2F6=1%2F3\"$ ....the other root\r
\n" ); document.write( "\n" ); document.write( "recall that product of roots: $\"c%2Fa\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"alpha%2A2alpha=+c%2Fa\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"2%28alpha%29%5E2=+c%2F2\"$
\n" ); document.write( " $\"%28alpha%29%5E2=+c%2F4\"$ substitute $\"1%2F6\"$ for $\"alpha\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"%281%2F6%29%5E2=+c%2F4\"$
\n" ); document.write( " $\"%281%2F36%29=+c%2F4\"$
\n" ); document.write( " $\"4%281%2F36%29=+c\"$
\n" ); document.write( " $\"cross%284%29%281%2Fcross%2836%299%29=+c\"$
\n" ); document.write( " $\"c=1%2F9\"$ \r
\n" ); document.write( "\n" ); document.write( "your equation is:
\n" ); document.write( " $\"2x%5E2-x%2B1%2F9=0\"$ \r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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