Let e_1 and e_2 be two distinct roots of the equation x^2-ax+2=0.

Question

Let ${e}_{1}$ and ${e}_{2}$ be two distinct roots of the equation ${x}^{2}-ax+2=0$. Let the sets ${a\in R:{e}_{1}$ and ${e}_{2}$ are the eccentricities of hyperbolas $}=(\alpha ,\beta )$, and ${a\in R:{e}_{1}$ and ${e}_{2}$ are the eccentricities of an ellipse and a hyperbola, respectively $}=(\gamma ,\infty)$.

Then ${\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}$ is equal to:

Accepted Answer

For the quadratic equation

$x^2-ax+2=0$

let its two distinct roots be $e_1$ and $e_2$.

We are given that these roots are to be interpreted as eccentricities.

Recall:

For a hyperbola, eccentricity $e>1$

For an ellipse, eccentricity satisfies $0

Also, since $e\_1,e\_2$ are roots of the quadratic, by Vieta's formulas,

$$ e_1+e_2=a,\qquad e_1e_2=2 $$

and since the roots are distinct real roots,

$$ a^2-8>0 \quad \Rightarrow \quad |a|>2\sqrt2 $$

Now we analyse the two sets.

1. Both $e\_1$ and $e\_2$ are eccentricities of hyperbolas
This means both roots must satisfy

$$ e_1>1,\qquad e_2>1 $$

Since

$e_1e_2=2>0$

the two roots have the same sign.

If both are hyperbola eccentricities, both must be positive and greater than $1$.

Now, for both roots of

$x^2-ax+2=0$

to be greater than $1$, we use the NCERT-style condition:

For both roots to be greater than $1$, we need

roots real and distinct:

$a^2-8>0$

sum positive:

$a>0$

value of polynomial at $x=1$ positive, because $1$ must lie to the left of both roots:

$$ f(1)=1-a+2=3-a>0 \text{ or }

Let us be careful.

Since the parabola opens upward, if both roots are greater than $1$, then $x=1$ lies outside the roots on the left side, so

$$ f(1)>0 $Hence,$ 3-a>0 \Rightarrow aAlso, for both roots to be greater than $1$, the axis of symmetry should be to the right of $1$:

$$ \frac a2>1 \Rightarrow a>2 $$

Combining:

$$ 2But also we need distinct real roots, and in this interval automatically $a>2\sqrt2$ is needed. Hence

$$ 2\sqrt2So,

$$ (\alpha,\beta)=(2\sqrt2,\,3) $$

Thus,

$$ \alpha=2\sqrt2,\qquad \beta=3 $$

2. $e\_1$ is eccentricity of an ellipse and $e\_2$ is eccentricity of a hyperbola, respectively
This means one root lies in $(0,1)$ and the other lies in $(1,\infty)$.

Since

$e_1e_2=2>0$

both roots are positive. So the only possibility is:

one root in $(0,1)$

the other root greater than $1$

Now let the roots be distinct and positive. Since their product is $2$, if one root is less than $1$, the other must automatically be greater than $2$, hence greater than $1$.

So we only need one root to be less than $1$.

For that, $x=1$ must lie between the two roots. Since the parabola opens upward, this happens when

$f(1)That is,$ 1-a+2$3-a$ a>3 $$

Also for real distinct roots, $a\^2-8>0$, which is already true for $a>3$.

Hence the set is

$$ (\gamma,\infty)=(3,\infty) $So,$ \gamma=3 $$

3. Find $\alpha\^2+\beta\^2+\gamma\^2$
Now,

$$ \alpha^2=(2\sqrt2)^2=8 $$ \beta^2=3^2=9 $$ \gamma^2=3^2=9 $Therefore,$ \alpha^2+\beta^2+\gamma^2=8+9+9=26 $So the correct answer is$ \boxed{26} $

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