Question

Whats the roots of the following three problems:

Answer 1

QUESTION 1

The given function is

$f(x)=\frac{(x-3)(x-1)}{(x-1)(x+2)}$

We simplify to get;

$f(x)=\frac{x-3}{x+2}$

This function is equal to zero when $x-3=0$

$\Rightarrow x=3$

QUESTION 2

The given function is

$f(x)=\frac{5x^2-10x+5}{2x^2-5x+3}$

$f(x)=\frac{5(x^2-2x+1)}{2x^2-5x+3}$

We factor to get;

$f(x)=\frac{5(x-1)^2}{(2x-3)(x-1)}$

$f(x)=\frac{5(x-1)}{(2x-3)}$

This function equals zero when

$5(x-1)=0$

$x=1$

QUESTION 3

The given function is

$f(x)=\frac{x^3-8}{x^2-6x+8}$

We factor to get,

$f(x)=\frac{(x-2)(x^2+2x+4)}{(x-2)(x+4)}$

$f(x)=\frac{x^2+2x+4}{x+4}$

The function equals zero when $x^2+2x+4=0$

$D=b^2-4ac$

$D=2^2-4(1)(4)$

$D=-12$

Hence the equation has no real roots.

The complex roots are

$x=-\sqrt{3}i-1$ or $x=-1+\sqrt{3}i$