If the following equation has no real roots, what are the possible values of p...

Step 1
We know that, the quadr. eqn,
${\displaystyle a{x}^2+bx+c=0}$ will have no real roots ${\displaystyle \iff b^2-4ac<0}$
Therefore, in our case, we must have,
${\displaystyle {(3p+1)}^2-4\times 1\times (-p)<0}$
${\displaystyle \therefore 9p^2+6p+1+4p<0}$, i.e., ${\displaystyle 9p^2+10p+1<0}$
${\displaystyle \therefore (p+1)(9p+1)<0}$
${\displaystyle \therefore \text{Case}\left(1\right):(p+1)<0}$ and ${\displaystyle (9p+1)>0}$, or,
Case ${\displaystyle \left(2\right):(p+1)>0}$, and ${\displaystyle (9p+1)<0}$
Case ${\displaystyle \left(1\right):(p+1)<0}$ and ${\displaystyle (9p+1)>0}$
${\displaystyle \therefore p<-1}$ and ${\displaystyle p\succ \frac{1}{9}}$. This is impossible.
Similarly, in Case (2), we have, ${\displaystyle p\succ 1,\mathrm{\&},pM-\frac{1}{9}}$, or
${\displaystyle -1<p<-\frac{1}{9}}$
We conclude that, ${\displaystyle \in (-1,-\frac{1}{9})}$

Step 1
If it has no real roots, then the discriminant of quadratic formula
${\displaystyle \mathrm{\Delta }={(3p+1)}^2+4p<0}$
${\displaystyle \mathrm{\Delta }=9p^2+6p+1+4p=9p^2+10p+1<0}$
Analyze this last expresion: represents a parabola for all values of p. Lets determine his roots
${\displaystyle p=\frac{-10\pm \sqrt{100-36}}{18}=\frac{-10\pm 8}{18}=-1}$ and ${\displaystyle -\frac{1}{9}}$
So in our case, the values of p for which the first equation has not real roots are in the interval ${\displaystyle (-1,-\frac{1}{9})}$

Step 1 If a quadratic equation has no real roots then its discriminant is less than zero. The discriminant of general quadratic equation $a{x}^2+bx+c=0$ is $b^2-4ac$ We have the equation $x^2+(3p+1)x-p=0$ and its discriminant is $(3p+1{)}^2-4\times 1\times (-p)=9p^2+6p+1+4p$ $=9p^2+10p+1=(9p+1)(p+1)$ Hence if $x^2+(3p+1)x-p=0$ has no real roots then $(9p+1)(p+1)<0$ Now there are two possiboloties 1) $9p+1<0$ and $p+1>0$ which means $p<-\frac{1}{9}$ and $p>-1$S. This is possible when $-1<p<-\frac{1}{9}$ 2) $9p+1>0$ and $p+1<0$ which means $p>-\frac{1}{9}$ and $p<-1$. But this is just not possible. Hence for $x^2+(3p+1)x-p=0$ to have no real roots we must have \(-1