Question

If the absolute value of the difference of roots of the equation ${{x}^{2}}+px+1=0$ exceeds $\sqrt{3p}$
(a) $p<-1$ or $p>4$
(b) $p>4$
(c) $-1 < p <4$
(d) $0\le p<4$

Answer 1

Hint: In this question, we first need to find the sum of the roots and product of the roots for the given equation on comparing it with $a{{x}^{2}}+bx+c=0$ given by the formula $\alpha +\beta =\dfrac{-b}{a}$ and $\alpha \beta =\dfrac{c}{a}$. Then, find the difference between the roots using the formula $x-a=\sqrt{{{\left( x+a \right)}^{2}}-4ax}$ and then substitute the respective value in the given condition $\left| \alpha -\beta \right|>\sqrt{3p}$ to get the result.

Complete step-by-step solution -
Now, from the given question we have the quadratic equation as
$\Rightarrow {{x}^{2}}+px+1=0$
Let us assume the roots of this quadratic equation as $\alpha ,\beta$
Relation between roots and coefficients:
If the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are $\alpha ,\beta$, then
Sum of the roots is given by the formula
$\Rightarrow \alpha +\beta =\dfrac{-b}{a}$
Product of the roots is given by the formula
$\Rightarrow \alpha \beta =\dfrac{c}{a}$
Now, on comparing the given equation with the standard form we get,
$a=1,b=p,c=1$
Now, let us find the sum of the roots
$\Rightarrow \alpha +\beta =\dfrac{-b}{a}$
Now, on substituting the respective values we get,
$\Rightarrow \alpha +\beta =-p$
Let us now find the product of the roots
$\Rightarrow \alpha \beta =\dfrac{c}{a}$
Now, on substituting the respective values we get,
$\Rightarrow \alpha \beta =1$
Now, from the given condition in the question we have,
$\Rightarrow \left| \alpha -\beta \right|>\sqrt{3p}$
Let us now find the difference between the roots
As we already know that from the factorisation of polynomials we have
$x-a=\sqrt{{{\left( x+a \right)}^{2}}-4ax}$
Now, from the above formula we get,
$\Rightarrow \alpha -\beta =\sqrt{{{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta }$
Now, on substituting the respective values we get,
$\Rightarrow \alpha -\beta =\sqrt{{{\left( -p \right)}^{2}}-4\times 1}$
Now, this can be further written as
$\Rightarrow \alpha -\beta =\sqrt{{{p}^{2}}-4}$
Let us now substitute this value in the condition we have
$\Rightarrow \left| \alpha -\beta \right|>\sqrt{3p}$
Now, on substituting the respective value we get,
$\Rightarrow \left| \sqrt{{{p}^{2}}-4} \right|>\sqrt{3p}$
Let us now do the squaring on both sides
$\Rightarrow {{p}^{2}}-4>3p$
Now, on rearranging the terms we get,
$\Rightarrow {{p}^{2}}-3p-4>0$
Now, this can be also written as
$\Rightarrow {{p}^{2}}+p-4p-4>0$
Now, on taking the common terms and writing it further we get,
$\Rightarrow p\left( p+1 \right)-4\left( p+1 \right)>0$
Now, this can be further written as
$\Rightarrow \left( p-4 \right)\left( p+1 \right)>0$
Now, the possibility is that either both should be greater than 0 or less than 0
Now, on considering the above condition we get
$\Rightarrow p-4>0,p+1>0$
$\Rightarrow p>4,p>-1$
Or
$\Rightarrow p-4<0,p+1<0$
$\Rightarrow p<4,p<-1$
Thus, the condition we get is
$\therefore p>4\text{ or }p<-1$
Hence, the correct option is (a).

Note: Instead of finding the difference of roots using sum and product of roots we can also solve it by finding the roots of the given equation using direct formula and then find their difference and simplify further to get the result. It is important to note that after squaring we do not mention absolute because as absolute means both positive and negative. But, when we square it changes to positive value so after squaring we don't use the absolute symbol.
It is also to be noted that while writing the inequality either both can be positive or both can be negative. So, it can be one condition from the first one or another condition from the second one.