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

Question

I wanted to find the values of (a) for which the function $f(x)=x-2+\frac{a-3}{x}$ has more than one root.

I know that the equation needs to be set equal to zero, from that step onward I have no idea how to proceed:

$$x-2+\frac{a-3}{x}=0$$ $$x^{2}-2x+a-3=0$$

from this point forward.

Answer 1

HINT: It may be rewritten as $(x-1)^2+a-4=0$ or $(x-1)^2=4-a$, hence $4-a$ is positive.

Answer 2

For a general quadratic equation $ax^2 + bx + c = 0$, the roots $\alpha_1$ and $\alpha_2$ are given by $$ \alpha_{1,2} = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ For there to be two roots, we'd need the discriminant $b^2 - 4ac$ to be non-zero, otherwise there'd only be one root that's $-b/2a$.

In other words, we desire $$ (-2)^2 - 4(1)(a-3) \neq 0 \iff a \neq 4 $$

As noted by Zain, if you're only looking for real roots, for $\sqrt{b^2 - 4ac}$ to be defined we desire that $$b^2 - 4ac > 0 \iff a < 4$$

Answer 3

$x-2+\frac{a-3}{x}=0$

$x\neq0$, so

$x^2-2x+a-3=0$

$\Delta= (2)^2-4(1)(a-3)=4-4a+12=16-4a$

We have two real roots when $\Delta > 0$, that is when

$16-4a>0$, that is when $a<4$

Answer 4

$$f(x)=x-2+\frac{a-3}{x}$$ $$\color{blue}{x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$$

Distinct real solutions iff $b^2-4ac\geq0$

You should get

$$\boxed{\color{red}{x_{1,2}=1\pm\sqrt{4-a}}}$$