Find the x-intercepts of these two rational funtions?

Question

(x^2-4)/(x^2-9) and (x^3-6x+5)/(x^2-3)

Answer 1

Finding an 'x-intercept' means finding the value(s) for x that make the function go to zero. Since the functions are fractions, they go to zero when the numerator is zero. For the first one
x²-4 = 0 => x = ±2.
You do the next one ☺


Doug

Answer 2

y = (x^2-4) (x^2 -9) develop using a^2- b^2 = (a-b) (a+b)

(x-2 ) (x+2) (x-3) (x+3)
In order to have y=0 , 4 solutions 2,-2,3,-3
4 intercepts -3, -2, 2, 3

z = (x^3 -6x +5) ( x^2 -3)

x^3 -6x +5 has x=1 as a root so you can divide by (x-1 )(x^2+x-5)

this can be rewritten (x-1)(x-1.8)(x+1.8)

and x^2 -3 if you calculate 3^0.5 = (x-1.73)x+1.73)

so z= (x-1)(x-1.8)(x+2.3.8)(x-1.73)x+1.73)

5 intercepts -1.8 , -1.73, 1, 1.73 , 2.3

Answer 3

FIRST PROBLEM
Factorize and solve

(x+2)(x-2)
------------ = 0
(x+3)(x-3)

x= 2
x=-2

SECOND PROBLEM

Factorizing and solving

(x-1) (x² +x -5)
--------------------=0
x² - 3

We should use the general formula to get the second term result

x= 1 is the first root

now for x² +x -5=0

x= -1 ± √21 / 2

x= (-1 + √21)/2 second root

x= (-1 - √21)/2 third root