If both the roots of the quadratic equation (2p + 1) x² + (3p + 2)x + (p+1) = 0 and (6p + 1)x² + 9px + 3p - 1 = 0 are common, then what is the value of p² +3p-1?

Given:

Quadratic equation 1: (2p + 1) x² + (3p + 2)x + (p+1) = 0

Quadratic equation 2: (6p + 1)x² + 9px + 3p - 1 = 0

Both quadratic equations have common roots.

Formula Used:

If two quadratic equations a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0 have both roots common, then \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).

Calculations:

Comparing the coefficients of the given quadratic equations:

a₁ = 2p + 1, b₁ = 3p + 2, c₁ = p + 1

a₂ = 6p + 1, b₂ = 9p, c₂ = 3p - 1

Since both roots are common, we have:

\(\frac{2p + 1}{6p + 1} = \frac{3p + 2}{9p} = \frac{p + 1}{3p - 1}\)

First, equate the first two ratios:

\(\frac{2p + 1}{6p + 1} = \frac{3p + 2}{9p}\)

⇒ 9p(2p + 1) = (3p + 2)(6p + 1)

⇒ 18p² + 9p = 18p² + 3p + 12p + 2

⇒ 18p² + 9p = 18p² + 15p + 2

⇒ 9p - 15p = 2

⇒ -6p = 2

⇒ p = -2/6 = -1/3

Now, let's verify if this value of p satisfies the equality of the second and third ratios:

\(\frac{3p + 2}{9p} = \frac{p + 1}{3p - 1}\)

Substitute p = -1/3:

\(\frac{3(-\frac{1}{3}) + 2}{9(-\frac{1}{3})} = \frac{-\frac{1}{3} + 1}{3(-\frac{1}{3}) - 1}\)

\(\frac{-1 + 2}{-3} = \frac{-\frac{1}{3} + \frac{3}{3}}{-1 - 1}\)

\(\frac{1}{-3} = \frac{\frac{2}{3}}{-2}\)

\(\frac{1}{-3} = \frac{2}{3 \times -2}\)

\(\frac{1}{-3} = \frac{2}{-6}\)

\(\frac{1}{-3} = \frac{1}{-3}\)

The value of p = -1/3 satisfies the condition for common roots.

Now, we need to find the value of p² + 3p - 1.

Substitute p = -1/3:

p² + 3p - 1 = \((-\frac{1}{3})^2 + 3(-\frac{1}{3}) - 1 \)

⇒ p² + 3p - 1 = \(\frac{1}{9} - 1 - 1 \)

⇒ p² + 3p - 1 = \(\frac{1}{9} - 2 \)

⇒ p² + 3p - 1 = \(\frac{1}{9} - \frac{18}{9}\)

⇒ p² + 3p - 1 = \(-\frac{17}{9}\)

∴ The value of p² + 3p - 1 is -17/9.