Question

How do you solve `3/(z^2-z-2)+18/(z^2-2z-3)=(z+21)/(z^2-z-2)` and check for extraneous solutions?

Answer 1

`z=-3`
Or
`z=6`

Explanation:

`3/(z^2-z-2)+18/(z^2-2z-3)=(z+21)/(z^2-z-2)`

`rArr3/(z^2-z-2)+18/(z^2-2z-3)-(z+21)/(z^2-z-2)=0`

To solve this equation we should find the common denominator,
so we have to factorize the denominators of the fractions above.

Let us factorize `color(blue)(z^2-z-2)` and `color(red)(z^2-2z-3)`

We can factorize using this method `X^2+color(brown)SX+color(brown)P`
where `color(brown)S` is the sum of two real number `a` and `b`
and
`color(brown)P` is their product
`X^2+color(brown)SX+color(brown)P=(X+a)(X+b)`

`color(blue)(z^2-z-2)`
Here,`color(brown)S=-1 and color(brown)P=-2`
so, `a=-2 and b=+1`
Thus,
`color(blue)(z^2-z-2=(z-2)(z+1)`

Factorize `color(red)(z^2-2z-3)`
Here,`color(brown)S=-2 and color(brown)P=-3`
so, `a=-3 and b=+1`
Thus,
`color(red)(z^2-2z-3=(z-3)(z+1)`

let us start solving the equation:

`3/color(blue)(z^2-z-2)+18/color(red)(z^2-2z-3)-(z+21)/color(blue)(z^2-z-2)=0`

`rArr3/color(blue)((z-2)(z+1))+18/color(red)((z-3)(z+1))-(z+21)/color(blue)((z-2)(z+1))=0`

`rArr(3(color(red)(z-3))+18(color(blue)(z-2))-(z+21)(color(red)(z-3)))/((z-2)(z-3)(z+1))=0`

`rArr(3z-9+18z-36-(z^2-3z+21z-63))/((z-2)(z-3)(z+1))=0`

`rArr(3z-9+18z-36-(z^2+18z-63))/((z-2)(z-3)(z+1))=0`

`rArr(3z-9+18z-36-z^2-18z+63)/((z-2)(z-3)(z+1))=0`

`rArr(3z-9cancel(+18z)-36-z^2cancel(-18z)+63)/((z-2)(z-3)(z+1))=0`

`rArr(-z^2+3z+18)/((z-2)(z-3)(z+1))=0`

As we know a fraction `color(orange)(m/n=0rArrm=0)`

`-z^2+3z+18=0`

`color(green)delta=(3)^2-4(-1)(18)=9+72=81`
Roots are:
`x_1=(-3+sqrt81)/(2(-1))=(-3+9)/(-2)=-3`
`x_1=(-3-sqrt81)/(2(-1))=(-3-9)/(-2)=6`

`-z^2+3z+18=0`
`(z+3)(z-6)=0`
`z+3=0rArrcolor(brown)(z=-3)`
Or
`z-6=0rArrcolor(brown)(z=6)`