Question

If the roots are real and equal, find the value of ‘p’: px² + (p =3) x + 1 = 0.

Answer 1

Solution:

If the roots of given Quadratic equation are real and equal than Its discriminate must be zero

$D = \sqrt{ {b}^{2} - 4ac} = 0$

Here in the Quadratic equation

$p {x}^{2} + (p + 3)x + 1 = 0 \\ \\ a = p \\ \\ b = p + 3 \\ \\ c = 1 \\ \\ so \\ \\ \sqrt{ {(p + 3)}^{2} - 4(p)(1) } = 0 \\ \\ ( {p + 3)}^{2} - 4p = 0 \\ \\ {p}^{2} + 9 + 6p - 4p = 0 \\ \\ {p}^{2} + 2p + 9 = 0$

it's a Quadratic equation again

$p_{1,2} = \frac{ - 2 ± \sqrt{4 - 18} }{2} \\ \\ p_{1,2} = \frac{ - 2 ± \sqrt{ - 12} }{2} \\ \\ p_{1,2} = \frac{- 2 + 2i \sqrt{3} }{2} \\ \\ p_{1}= - 1 + i \sqrt{3} \\ \\ p_{2} = - 1 - i \sqrt{3}$

will be the value of p.

No such real value of p exist.

*I guess it's p+3,so Solution is based on that value.