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

May 8, 2018

$- 1 < p < - \frac{1}{9} , \mathmr{and} , p \in \left(- 1 , - \frac{1}{9}\right)$.

#### Explanation:

We know that, the quadr. eqn.,

$a {x}^{2} + b x + c = 0 \text{ will have no real roots } \Leftrightarrow {b}^{2} - 4 a c < 0.$

Therefore, in our case, we must have,

${\left(3 p + 1\right)}^{2} - 4 \cdot 1 \cdot \left(- p\right) < 0$.

$\therefore 9 {p}^{2} + 6 p + 1 + 4 p < 0 , i . e . , 9 {p}^{2} + 10 p + 1 < 0$.

$\therefore \left(p + 1\right) \left(9 p + 1\right) < 0$.

$\therefore \text{ Case } \left(1\right) : \left(p + 1\right) < 0 \mathmr{and} \left(9 p + 1\right) > 0 , \mathmr{and} ,$

$\text{ Case } \left(2\right) : \left(p + 1\right) > 0 , \mathmr{and} \left(9 p + 1\right) < 0$.

Case (1) : $\left(p + 1\right) < 0 \mathmr{and} \left(9 p + 1\right) > 0$.

$\therefore p < - 1 \mathmr{and} p > - \frac{1}{9}$. This is impossible.

Similarly, in Case (2) , we have, p gt -1, &, p lt -1/9, or,

$- 1 < p < - \frac{1}{9}$.

We conclude that, $p \in \left(- 1 , - \frac{1}{9}\right)$.

Enjoy Maths.!

May 8, 2018

If it has no real roots, then the discriminant of quadratic formula $\Delta = {\left(3 p + 1\right)}^{2} + 4 p < 0$

$\Delta = 9 {p}^{2} + 6 p + + 1 + 4 p = 9 {p}^{2} + 10 p + 1 < 0$

Analyze this last expresion: represents a parabola for all values of p. Lets determine his roots

$p = \frac{- 10 \pm \sqrt{100 - 36}}{18} = \frac{- 10 \pm 8}{18} = - 1 \mathmr{and} - \frac{1}{9}$

So in our case, the values of p for which the first equation has not real roots are in the interval $\left(- 1 , - \frac{1}{9}\right)$

May 8, 2018

$- 1 < p < - \frac{1}{9}$

#### Explanation:

If a quadratic equation has no real roots then its discriminant is less than zero. The discriminant of general quadratic equation $a {x}^{2} + b x + c = 0$ is ${b}^{2} - 4 a c$.

We have the equation ${x}^{2} + \left(3 p + 1\right) x - p = 0$ and its discriminant is

${\left(3 p + 1\right)}^{2} - 4 \cdot 1 \cdot \left(- p\right) = 9 {p}^{2} + 6 p + 1 + 4 p$

= $9 {p}^{2} + 10 p + 1 = \left(9 p + 1\right) \left(p + 1\right)$

Hence if ${x}^{2} + \left(3 p + 1\right) x - p = 0$ has no real roots then

$\left(9 p + 1\right) \left(p + 1\right) < 0$

Now there are two possibilities

(1) $9 p + 1 < 0$ and $p + 1 > 0$ which means $p < - \frac{1}{9}$ and $p > - 1$. This is possible when $- 1 < p < - \frac{1}{9}$

(2) $9 p + 1 > 0$ and $p + 1 < 0$ which means $p > - \frac{1}{9}$ and $p < - 1$. But this is just not possible.

Hence for ${x}^{2} + \left(3 p + 1\right) x - p = 0$ to have no real roots

we must have $- 1 < p < - \frac{1}{9}$.