Question

The equation ${{x}^{2}}+ax-{{a}^{2}}-1=0$ will have roots of opposite signs if:

A) $a\in \left( -\infty ,\infty \right)$

B) $a\in \left[ -1,1 \right]$

C) $a\in \left( -\infty ,-1 \right)\cup \left( 1,\infty \right)$

D) None of these

Answer 1

Hint: Here we will use properties of roots of the quadratic equation. If the roots are of the opposite sign then their product will be negative. And product = $\dfrac{\text{constant}}{\text{coefficient of }{{x}^{2}}}$ So, Product is less than 0.

Complete step-by-step solution:

Given: ${{x}^{2}}+ax-{{a}^{2}}-1=0$ will have roots of the opposite signs.

${{x}^{2}}+ax-({{a}^{2}}+1)=0$ as roots have opposite signs,

Compare the given equation with the standard form-

$ a{{x}^{2}}+bx+c=0 $

$  \therefore a=1 $

$ \therefore b=a $

$ \therefore c=-({{a}^{2}}+1) $

Product of roots is less than zero by property -

$\Rightarrow \alpha \times \beta <0\text{ and }D>0$

$\Rightarrow \dfrac{c}{a}<0\text{ }and\text{ }{{b}^{2}}-4ac>0$

Put values in the above conditions -

$\Rightarrow \dfrac{-\left( {{a}^{2}}+1 \right)}{1}<0\text{ }and\text{ }{{\text{a}}^{2}}-4(1)\{-({{a}^{2}}+1)\}>0\text{ }$

By simplification -

$\Rightarrow \left( {{a}^{2}}+1 \right)>0\text{ and }{{\text{a}}^{2}}+4{{a}^{2}}+4>0$

$\Rightarrow a\in R\text{ and 5}{{\text{a}}^{2}}+4>0$

$Now,\text{ a}\in \text{R and }{{\text{a}}^{2}}>-\dfrac{4}{5}$

So, in both cases, $a\in R$ which means $a\in \left( -\infty ,\infty \right)$

Hence option (B) is the correct answer.

Note: In these types of problems where nature of root and equation is given. We first try to connect the given nature of roots to coefficients and product or sum of roots whichever is needed. Then check inequality after assigning the required values. Also, here we have written as if signs of roots are opposite then the product of roots will be negative, as with the same sign product of roots will be positive. As per the property it is like (-)(-)=(+) and (+)(+)=(+).