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) $(3,\infty )$
B) $(2,\infty )$
C) $(1,\infty )$
D) $(-\infty ,1)\cup (1,2)$

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,
$\Rightarrow \sqrt{8} < a < \infty$
For, $a<-\sqrt{8}$,
It ranges from,
$\Rightarrow - \infty < a<- \sqrt{8} $
We can write, $\sqrt{8}=2\sqrt{2}$.
Now let us check the option.
From $(3,\infty )$ not satisfied.
$(2,\infty )$ can satisfy the condition.
Also, $(1,\infty )$ do not satisfy the condition.
After that, $(-\infty ,1)\cup (1,2)$ do not satisfy the condition.
So, $a$ belongs to $(2,\infty )$.

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.