This is an MCQ-based quiz for GRE on **How To Find Excluded Values**.

This includes Excluded values which are values that will make the denominator of a fraction equal to 0. You can't divide by 0, so it's very important to find these excluded values when you're solving a rational expression.

If f(x)=(x^2−25) / (3x^2+4x−7) then which values of x cannot exist?

x=1

x= -1

x=5

x= -5

x= -7

Which of the following provides the complete solution set for |2 - z| > 0 ?

z<2

No solutions

z=2

z!=2

z>2

If f(x) = (x^2 - 16)^1/2 then which cannot be an x value?

x=3.5

x= -6.7

x=16

x=4

x= -5.2

Find the excluded values of the following algebraic fraction

(2x^2−5x−12) / (x^2−11x+28)

x!= -11

x!=4

x!=4,7

The numerator cancels all the binomials in the denomniator so ther are no excluded values.

x!=7