What are the roots of the function f(x)= (x^2-5x+4)/(x-4) ?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The roots of a function are the values of x at which f(x) = 0

f(x) = (x^2 - 5x + 4)/(x - 4) = 0

=> (x^2 - 5x + 4)/(x - 4) = 0

=> (x^2 - 4x - x + 4)/(x - 4) = 0

=> (x(x - 4) - 1(x - 4))/(x - 4) = 0

=> (x - 1)(x - 4)/(x - 4) = 0

=> (x - 1) = 0

=> x = 1

We get x = 1. At x = 4 the function is not defined.

The root of the function is x = 1.

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

f(x) = (x^2 - 5x +4) / (x-4)

We need to find the roots of the equation.

First we know that the domain is R- { 4}.

Now we know that the roots of f(x) are the zeros of the numerator,

==> x^2 - 5x + 4= 0

We will factor.

==> (x-4)(x-1) = 0

==> x = 4

==> x = 1

But 4 does not belong to the domain of f(x).

Then the only root of f(x) is x= 1.

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