document.write( "Question 845307: My textbook for Algebra asks for us to \"Solve the equation\" but our teacher didn't go over this and I'm confused. The problem in question is $\"+1%2Fx%5E2+%2B+8%2Fx+%2B15=0\"$ am I supposed to use the quadratic formula for this? Thanks for your trouble! \n" ); document.write( "

Algebra.Com's Answer #509225 by KMST(5328) $\"\"$ $\"About$

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You could go about it one of two ways:
\n" ); document.write( "1) You could multiply both sides of the equation times $\"x%5E2\"$ to get
\n" ); document.write( " $\"1%2B8x%2B15x%5E2=0\"$ <--> $\"15x%5E2%2B8x%2B1=0\"$ and solve that equation.
\n" ); document.write( "(That is probably the expected way).
\n" ); document.write( "2) You could say $\"y=1%2Fx\"$ , and that makes $\"y%5E2=1%2Fx%5E2\"$ ,
\n" ); document.write( "and substituting you would get
\n" ); document.write( " $\"y%5E2%2B8y%2B15=0\"$ .
\n" ); document.write( "Solving that equation you get solutions for $\"y\"$ ,
\n" ); document.write( "which you can substitute into $\"y=1%2Fx\"$ to find the values for $\"x\"$ .
\n" ); document.write( "
\n" ); document.write( "MULTIPLYING TIMES $\"x%5E2\"$ :
\n" ); document.write( "We know that $\"x%5E2\"$ is not zero, because it is in a denominator and if it were zero, $\"1%2Fx%5E2\"$ would not exist. If we found $\"x=0\"$ as a solution, we would discard it, because we know it is not a valid solution of
\n" ); document.write( " $\"+1%2Fx%5E2+%2B+8%2Fx+%2B15=0\"$ .
\n" ); document.write( "So we can safely multiply times $\"x%5E2\"$ both sides of the equal sign.will get
\n" ); document.write( " $\"+1%2Fx%5E2+%2B+8%2Fx+%2B15=0\"$
\n" ); document.write( " $\"x%5E2%2A%281%2Fx%5E2+%2B+8%2Fx+%2B15%29=x%5E2%2A0\"$
\n" ); document.write( " $\"+x%5E2%281%2Fx%5E2%29+%2B+x%5E2%2A%288%2Fx%29+%2Bx%5E2%2A15=0\"$
\n" ); document.write( " $\"1%2B8x%2B15x%5E2=0\"$
\n" ); document.write( " $\"15x%5E2%2B8x%2B1=0\"$
\n" ); document.write( "The quadratic formula says that the solutions to
\n" ); document.write( " $\"ax%5E2%2Bbx%2Bc=0\"$ are given by
\n" ); document.write( " $\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"$
\n" ); document.write( " $\"15x%5E2%2B8x%2B1=0\"$ is $\"ax%5E2%2Bbx%2Bc=0\"$ with
\n" ); document.write( " $\"a=15\"$ , $\"b=8\"$ , and $\"c=1\"$ .
\n" ); document.write( "So $\"x+=+%28-8+%2B-+sqrt%288%5E2-4%2A15%2A1+%29%29%2F%282%2A15%29+\"$
\n" ); document.write( " $\"x+=+%28-8+%2B-+sqrt%2864-60%29%29%2F30+\"$
\n" ); document.write( " $\"x+=+%28-8+%2B-+sqrt%284%29%29%2F30+\"$
\n" ); document.write( " $\"x+=+%28-8+%2B-+2%29%2F30+\"$
\n" ); document.write( "The solutions are
\n" ); document.write( " $\"x+=+%28-8+%2B+2%29%2F30+=-6%2F30=highlight%28-1%2F5%29\"$ and $\"x+=+%28-8+-+2%29%2F30+=-10%2F30=highlight%28-1%2F3%29\"$
\n" ); document.write( "
\n" ); document.write( "NOTE:
\n" ); document.write( "You could use the quadratic formula, or you could solve by factoring, or you could even solve by completing the square.
\n" ); document.write( "For $\"15x%5E2%2B8x%2B1=0\"$ , using the quadratic formula is very reasonable,
\n" ); document.write( "because the other options do not seem any easier.
\n" ); document.write( "If you had changed variables (option 2) above) to end with
\n" ); document.write( " $\"y%5E2%2B8y%2B15=0\"$ ,
\n" ); document.write( "you may find that factoring is the easiest option.
\n" ); document.write( "You may immediately see that you can factor to get
\n" ); document.write( " $\"%28y%2B3%29%28y%2B5%29=0\"$ ,
\n" ); document.write( "which would tell you that the solutions are $\"y=-3\"$ and $\"y=-5\"$ .
\n" ); document.write( " $\"y=-3\"$ means $\"1%2Fx=-3\"$ --> $\"1=-3x\"$ --> $\"1%2F%28-3%29=x\"$ --> $\"x=-1%2F3\"$
\n" ); document.write( " $\"y=-5\"$ means $\"1%2Fx=-5\"$ --> $\"1=-5x\"$ --> $\"1%2F%28-5%29=x\"$ --> $\"x=-1%2F5\"$ \n" ); document.write( "

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