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 .

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:

First, equate the first two ratios:

⇒ 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:

Substitute p = -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 =

⇒ p² + 3p - 1 =

⇒ p² + 3p - 1 =

⇒ p² + 3p - 1 =

⇒ p² + 3p - 1 =

∴ The value of p² + 3p - 1 is -17/9.