How do you simplify and find the restrictions for #(x^2+3x-18)/(x^2-36)#?

3 Answers
Nov 16, 2017

Answer:

The restricted values of #x# are: #x_1=6# and #x_2=-6#.

Explanation:

There is no simplification possible.

To find the restrictions you have to see for what values of #x# there's no solution. That would be when the denominatior is #0#.

So you get,
#x^2-36=0#

you isolate #x#,
#x^2=36#

and you do the square root in both sides,
#x=+-sqrt(36)=+-6#

so these are the restricted values on x:
#x_1=6#
#x_2=-6#

Nov 16, 2017

#f(x)=##(x^2+3x-18)/(x^2-36)=((x-3)(x+6))/((x-6)(x+6))=(x-3)/(x-6)#

( #f(x)=0##<=>##x=3#

  • #x≠6 and x≠-6#
    #Df=(-oo,-6)U(-6,6)U(6,+oo)#
    #Df= R-{-6,6}# )
Nov 16, 2017

Answer:

#(x^2+3x-18)/(x^2-36)# simplifies to #(x-3)/(x-6)# with the restriction that #x!=6# and #x!=+6#

Explanation:

#(x^2+3x-18)/(x^2-36)#

#color(white)("XXX")=((x+6)(x-3))/((x+6)(x-6))#

#color(white)("XXX")#Note the division is only defined if #x!=+-6#

#color(white)("XXX")=(x-3)/(x-6)# provided #(x+6)!=0#