Question

If $e_1$ and $e_2$ are the roots of the equation $x^2-ax+2=0$, where $e_1$ and $e_2$ are the eccentricities of an ellipse and hyperbola, respectively, then the value of $a$ belongs to
A) $(3,\mathrm{\infty })$
B) $(2,\mathrm{\infty })$
C) $(1,\mathrm{\infty })$
D) $(-\mathrm{\infty },1)\cup (1,2)$

Answer 1

Hint:
We are given that, $e_1$ and $e_2$ are the roots of the equation $x^2-ax+2=0$, where $e_1$ and $e_2$ are the eccentricities of an ellipse and hyperbola. So, here,$e_1$ and $e_2$ are real. Since, it is real $b^2-4ac>0$. After that, apply the condition and solve it. Try it, you will get the range of $a$.

Complete step by step solution:
Now it is given that, $e_1$ and $e_2$ are the roots of the equation $x^2-ax+2=0$, where $e_1$ and $e_2$ are the eccentricities of an ellipse and hyperbola.
$\Rightarrow x^2-ax+2=0$
Here, $e_1$ and $e_2$ will be real.
So, $\Rightarrow b^2-4ac>0$ …………. (1)
Now comparing $x^2-ax+2$ with $a{x}^2+bx+c$ we get,
$\Rightarrow a=1,$ $b=-a$, $c=2$
We get,
$\Rightarrow b^2-4ac={(-a)}^2-4(1)(2)$
Simplifying we get,
$\Rightarrow b^2-4ac=a^2-8$ ……… (2)
 From (1) and (2), we get,
$\Rightarrow a^2-8>0$
Now adding four on both sides we get,
$\Rightarrow a^2-8+8>8$
Again, simplifying we get,
$\Rightarrow a^2>8$
Now taking square root we get,
$\Rightarrow a>\sqrt{8}$ and $a<-\sqrt{8}$
Now taking, $a>\sqrt{8}$,
So, it ranges from,
$\Rightarrow \sqrt{8}<a<\mathrm{\infty }$
For, $a<-\sqrt{8}$,
It ranges from,
$\Rightarrow -\mathrm{\infty }<a<-\sqrt{8}$
We can write, $\sqrt{8}=2\sqrt{2}$.
Now let us check the option.
From $(3,\mathrm{\infty })$ not satisfied.
$(2,\mathrm{\infty })$ can satisfy the condition.
Also, $(1,\mathrm{\infty })$ do not satisfy the condition.
After that, $(-\mathrm{\infty },1)\cup (1,2)$ do not satisfy the condition.
So, $a$ belongs to $(2,\mathrm{\infty })$.

The correct answer is option (B).

Additional information:
Quadratic Formula helps to evaluate the solution of quadratic equations replacing the factorization method. A quadratic equation is of the form of $a{x}^2+bx+c=0$, where $a,b$ and $c$ are real numbers, also called “numeric coefficients”. We know that a second-degree polynomial will have at most two zeros. Therefore, a quadratic equation will have at most two roots. By splitting the middle term, we can factorize quadratic polynomials.

Note:
1) The term $b^2-4ac$ in the quadratic formula is known as the discriminant of a quadratic equation. The discriminant of a quadratic equation reveals the nature of roots.
2) If the value of discriminant $=0$ i.e. $b^2-4ac=0$ the quadratic equation will have equal roots.
3) If the value of discriminant $<0$ i.e. $b^2-4ac<0$ the quadratic equation will have imaginary roots.
4) If the value of discriminant $>0$ i.e. $b^2-4ac>0$ then the quadratic equation will have real roots.