Caleb Snyder

2022-01-23

If the following equation has no real roots, what are the possible values of p for ${x}^{2}+\left(3p+1\right)x-p=0?$

Palandriy0

Expert

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

gekraamdbk

Expert

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

RizerMix

Expert

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 ${x}^{2}+\left(3p+1\right)x-p=0$ and its discriminant is $\left(3p+1{\right)}^{2}-4×1×\left(-p\right)=9{p}^{2}+6p+1+4p$ $=9{p}^{2}+10p+1=\left(9p+1\right)\left(p+1\right)$ Hence if ${x}^{2}+\left(3p+1\right)x-p=0$ has no real roots then $\left(9p+1\right)\left(p+1\right)<0$ 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 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 ${x}^{2}+\left(3p+1\right)x-p=0$ to have no real roots we must have \(-1