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?

1 Answer
Oct 14, 2016

Answer:

#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)#