Question

Pre Calc Help?? Much appreciated

Answer 1

`(18x^2-68x+24)/(x(x-2)(x-4))=\frac ax+\frac b{x-2}+\frac c{x-4}`

`18x^2-68x+24=a(x-2)(x-4)+bx(x-4)+cx(x-2)`

`x=0\implies24=8a\implies a=3`

`x=2\implies-40=-4b\implies b=10`

`x=4\implies40=8c\implies c=5`

So we have

`(18x^2-68x+24)/(x(x-2)(x-4))=\frac3x+(10)/(x-2)+\frac5{x-4}`

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`xy=12\implies y=\frac{12}x`

`x^2+\left(\frac{12}x\right)^2=40`

`x^2+(144)/(x^2)=40`

`x^4-40x^2+144=0`

`(x^2-36)(x^2-4)=0`

`x^2-36=0\implies x=\pm6`

`x^2-4=0\implies x=\pm2`

The solutions are then

`x=-6\implies y=-2`

`x=-2\implies y=-6`

`x=2\implies y=6`

`x=6\implies y=2`