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?

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OSSC Excise SI (Mains) Official Paper (Held On: 17 Oct, 2024 Shift 1)
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Answer (Detailed Solution Below)

Option 1 :

Detailed Solution

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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.

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