Question

At what values of x is f(x)=(x^2-2x-3)/(x-2)=0 ?

Answer 1

f(x)=(x^2-2x-3)/(x-2)=0 when x^2-2x-3=0
so D=b^2-4ac=4-4*1*(-3)=16
x1=(-b+sqrD)/2a==(2+4)2=3
x2=(-b-sqrD)/2a=(2-4)/2=-2

Answer 2

\[f(x) = \frac{x^2-2x-3}{x-2}=0\]
When \[x^2-2x-3=0\] (Denominator can't make that equation 0; only numerator can.)

Factor that polynomial, and you will get:\[(x-3)(x+1)=0\]
And its zeros are therefore: \[x = 3, -1\]

If you try to plug in 3, you will get:\[\frac{3^2-2(3)-3}{3-2}=0\]
Do the same for -1: \[\frac{(-1)^2-2(-1)-3}{3-(-1)}=0\]

Do be mindful that if a solution makes the denominator 0, then that solution yields an undefined value.

Answer 3

yea -1 sorry not -2 (typing mistake) :(