Question

If $\alpha$, $\beta$ are the roots of the quadratic equation $(p^2+p+1)x^2+(p-1)x+p^2=0$ such that unity lies between the roots, then the set of values of $p$ is

• A. $\varphi$
• B. $p\in (-\infty ,-1)\cup (0,\infty )$
• C. $p\in (-1,0)$
• D. $(-1,1)$

Answer 1

Answer: (C)

$p\in (-1,0)$

Quadratic equation : $(p^2+p+1)x^2+(p-1)x+p^2=0$
Comparing with the standard form of quadratic equation $a{x}^2+bx+c=0$,we get
$a=p^2+p+1,b=p-1andc=p^2$
Unity lies between the two roots $\Rightarrow$ Atleast one of the roots is positive
$a=p^2+p+1={(p+\frac{1}{2})}^2+\frac{3}{4}$
Hence,$a>0$
Product of roots = $\frac{c}{a}$ = $\frac{p^2}{p^2+p+1}$
Hence,product of roots is also positive as both numerator and denominator are positive $\Rightarrow$ the other smaller root is also positive
Since both roots are positive,their sum must be positive.
Hence,sum of roots = $\frac{-b}{a}$ = $\frac{-(p-1)}{(p^2+p+1)}$ = $\frac{1-p}{(p^2+p+1)}$ $>0$
$\Rightarrow$ $1-p>0$ (as denominator is always positive)
$\Rightarrow$ $p<1$ .................... (1)
Since $a=p^2+p+1>0$, the function is always positive except for the x values lying between the two roots.
Since unity lies between the two roots,the function value at x=1 should be negative
$\Rightarrow$ $(p^2+p+1){\left(1\right)}^2+(p-1)\left(1\right)+p^2<0$
$\Rightarrow$ $2p^2+2p<0$
$\Rightarrow$ $p(p+1)<0$
$\Rightarrow$ $p\in (-1,0)$ ................ (2)
From equation (1) and (2),we get $p\in (-1,0)$
Hence,option (C) is the correct choice.