Consider the equation : $$(a^2+a+1)x^2-2ax-a^2-1=0$$ Prove that for all values of $a$, the equation is quadratic.
So I know that the coefficient of $x^2$ should not be equal to 0
$$a^2+a+1≠ 0$$ but I have no idea how to work from there.
Without calculating the discriminant, prove that the equation (E) has two roots in R.
How do I do that ? I'm used to just calculating the discriminant and proving that it's positive which means (E) has 2 roots. What other way is there?
On
In order to see if $a^2+a+1$ equals to $0$ you could try to solve: $$a^2+a+1=0\implies a_{1,2}=\frac{-1\pm\sqrt{1^2-4}}{2}=\frac{-1\pm\sqrt{-3}}{2}.$$
As we have $\sqrt{-3}$, we can conclude that the quadratic equation has no real roots (is not zero for $a \in \mathbb{R}$).
If you fear using the solution for the quadratic equation you can also observe that the discriminant $b^2-4ac<0$ for $ax^2+bx+c$, hence the equation has no real solution.
On
In a little roundabout way:
0)$y(a) = a^3 $ is strictly monotonic in $\mathbb{R}$,
with a single, unique zero at $a=0$.
1)$y(a)=a^3-1$ is strictly monotonic $ \mathbb{R} $,
with a single, unique zero at $a=1.$
$y(a)= a^3-1=$
$ (a-1)(a^2+a+1) $.
$\rightarrow:$
$(a^2+a+1) \ne 0$ for $a \in \mathbb{R}$.
Hint: $$a^2+a+1 = a^2 + 2(a)(1/2)+1/4+3/4 = (a+1/2)^2+3/4$$