If $\alpha$ is the root(having least absolute value) of the equation$x^2-ax-1=0(a\in R^+)$

Question

If $\alpha$ is the root(having least absolute value) of the equation$x^2-ax-1=0(a\in R^+)$,then which of the following relation is correct?
$(A)\alpha<-1\hspace{1cm}(B)-1<\alpha<0\hspace{1cm}(C)0<\alpha<1\hspace{1cm}(D)\alpha>1$

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Let $\alpha,\beta$ are the roots of the equation $x^2-ax-1=0(a\in R^+)$ with $\alpha<\beta$

then $\alpha+\beta=a,\alpha\beta=-1$
How can i select the correct answer from this?

Answer 1

(0) Let $y = f(x) = x^2-ax-1 $.

(1) From $((x^2)) = 1 \gt 0$, what is the direction of the opening of the curve?

(2) From $\alpha\beta=-1$, can $\alpha$ and $\beta$ have the same sign?

(2.1) Further, if $\beta \gt \alpha$, which root must be negative?

(3) Where is the axis of symmetry? Will it lie on the right side of the y-axis?

Base on the above info, make a rough sketch of the graph.

(4) Show that $f(–1) > 0$.

(5) From (4), determine whether $\alpha$ can be placed on the left of x = –1 or not.