If \(Px^{2} + (P+1)x + P = 0\) has equal roots, find the values of P.
Answer Details
Since the quadratic equation has equal roots, the discriminant is equal to zero. Therefore, we have:
$$(P+1)^2 - 4P^2 = 0$$
Simplifying the equation above, we get:
$$P^2 + 2P + 1 - 4P^2 = 0$$
$$-3P^2 + 2P + 1 = 0$$
We can now solve for P using the quadratic formula:
$$P = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $a=-3$, $b=2$, and $c=1$.
Substituting these values, we get:
$$P = \frac{-2 \pm \sqrt{2^2 - 4(-3)(1)}}{2(-3)}$$
Simplifying, we have:
$$P = \frac{-2 \pm \sqrt{16}}{-6}$$
Therefore, the solutions are:
$$P_1 = \frac{-2 + 4}{-6} = \frac{-1}{3}$$
$$P_2 = \frac{-2 - 4}{-6} = 1$$
Hence, the values of P are $\frac{-1}{3}$ and 1. Therefore, the correct option is $\text{: 1 and }\frac{-1}{3}$.