Finding the Value of p for a Quadratic Equation with One Root Twice the Other

Given the quadratic equation (3x^2 px - 54 0), if one root is twice the other, how can we determine the value of (p)? This problem utilizes the fundamental concepts of quadratic equations and the relationships between their roots and coefficients. We will walk through the steps to solve for (p) by employing Vieta's formulas and algebraic manipulation.

Understanding the Problem

We are given a quadratic equation:

[[3x^2 px - 54 0]

Let's denote the roots by (r) and (2r). The key is to use Vieta's formulas, which relate the sum and product of the roots to the coefficients of the polynomial.

Using Vieta's Formulas

According to Vieta's formulas:

The sum of the roots (r 2r 3r) is equal to (-frac{p}{3}). The product of the roots (r cdot 2r 2r^2) is equal to (frac{54}{3} 18).

Solving for (r)

First, we solve the product of the roots:

[2r^2 18]

Dividing both sides by 2, we get:

[r^2 9]

Taking the square root of both sides, we obtain:

[r 3 quad text{or} quad r -3]

Calculating the Sum of the Roots

Next, we calculate the sum of the roots for both possible values of (r):

When (r 3), the roots are 3 and 6. The sum is (3 6 9). When (r -3), the roots are -3 and -6. The sum is (-3 (-6) -9).

Using Vieta's first formula, the sum of the roots (3r) is equal to (-frac{p}{3}):

For (r 3): (3r 3 cdot 3 9), so (-frac{p}{3} 9). Multiplying both sides by -3 gives (p -27). For (r -3): (3r 3 cdot (-3) -9), so (-frac{p}{3} -9). Multiplying both sides by -3 gives (p 27).

Therefore, the possible values for (p) are (-27) and (27). However, in typical contexts, positive values for coefficients are more relevant:

[boxed{-27}]

Additional Scenarios and Solutions

Let's explore a couple of additional scenarios to reinforce the process:

Scenario 1: Roots are (a) and (2a)

Given the equation (frac{54}{3} 18 2a^2), we solve for (a):

[[2a^2 18]

Dividing both sides by 2, we get:

[[a^2 9]

Taking the square root of both sides, we obtain:

[[a pm 3]

So, the roots are (pm 3) and (pm 6). Calculating the sum of the roots:

[[2a^2 18 implies a pm 3]

Using Vieta’s formula, (-frac{p}{3} 36 implies p -27)

Similarly, (-frac{p}{3} -36 implies p 27)

Scenario 2: Roots are (u) and (2u)

Given that (3u -frac{p}{3}) and (2u^2 frac{54}{3} 18), we solve for (u):

[[2u^2 18 implies u^2 9 implies u pm 3]

Therefore, (p -9u pm 27)

After checking, (p -27) gives the correct roots (x' -3, x'' -6).

Conclusion

In summary, the value of (p) for the quadratic equation (3x^2 px - 54 0) with one root being twice the other can be determined as (-27) through careful algebraic manipulation and Vieta's formulas.

Note: The context typically implies positive values for coefficients, so the most relevant solution is (-27).