Can someone explain the solution to "If $x^2+ax−3x−(a+2)=0$ has real and distinct roots, then minimum value of $(a^2+1)/(a^2+2)$ is?"

Question

Could we not have directly skipped to the last line of the solution? The conclusion seems disjointed from the previous lines.

[Question]:

If $x^2+ax−3x−(a+2)=0$ has real and distinct roots, then minimum value of $(a^2+1)/(a^2+2)$ is...?

[Given Solution]:

$x^2+ax−3x−(a+2)=0$

$D=(a−3)^2+4(a+2)$

$⇒D=a^2−2a+17$

$D1​=4−4(17)<0$

Therefore, $a^2−2a+17>0$ for all $a∈R$

Now, $(a^2+1)/(a^2+2)​=1−1/(a^2+2)​>1−(1/2)​=1/2 $

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What exactly were we able to prove in those first 5 lines of the solution? At first I thought we had proven that 'a' would always be real and never complex, but that doesn't seem to be the case.

Answer 1

For $x^2+ax−3x−(a+2)=0$ to have real and distinct roots, its discriminant $D$ must be strictly +ve. That is, we must have, $D\gt 0$, which is true because as you have rightly shown
$D=a^2−2a+17$ and since $D'=(2)^2-4(17)\lt 0$, We know that $ a^2-2a+17=0\implies$ the equation $D=0$ can not have any real root, that is $D\ne 0$ for any real $a$. In fact, $D=a^2-2a+17=(a-1)^2+16\gt 0 \;\;\forall a\in\mathbb R$. Therefore $a$ has to be real so that $D\gt 0$.

PS: 1) Geometrically, $D=a^2−2a+17$ represents a parabola on $D-a$ axis and since $D\gt 0$, it never intersects axis of $a$ and hence $D\ne 0$ rather $D\gt 0$.
2) In the first five lines, we showed that $a$ is real.

Answer 2

I think you are right, there are useless steps.

The minimum value of

$$\frac{a^2+1}{a^2+2}=1-\frac1{a^2+2}$$ is achieved by $a=0$, if this is an admissible value.

And indeed, $$\Delta(0)=0^2-2\cdot0+17>0$$

and we are done.

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In fact the author was solving the inequation $a^2-2a+17>0$ to find the smallest admissible value of $a^2$ (corresponding to the smallest value of $|a|$), and concluded that $a\in\mathbb R$.

Answer 3

In the equation $$x^2+ax−3x−(a+2)=0$$ is $-(a+2)$ the product of roots. It is given that the roots are real, therefore $$a \;\text{is real.}$$ The answer to your question is YES: the first steps of the given solution can be skipped, we could go directly to the last line.