Math Problem Statement

  1. For the equation 3x2 + px + 3 = 0 , p>0, if one of the root is square of the other, then p is equal to

Solution

To solve for pp in the equation 3x2+px+3=03x^2 + px + 3 = 0, where one root is the square of the other, we can proceed step by step.

Step 1: Let the roots be α\alpha and β\beta.

From the problem, we are given that one root is the square of the other. Let’s assume that α=β2\alpha = \beta^2 (without loss of generality).

Step 2: Use Vieta's formulas.

For a quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0, Vieta’s formulas give us the relationships between the coefficients and the roots:

  • Sum of the roots: α+β=ba\alpha + \beta = -\frac{b}{a}
  • Product of the roots: αβ=ca\alpha \beta = \frac{c}{a}

In this case, the equation is 3x2+px+3=03x^2 + px + 3 = 0, so:

  • a=3a = 3, b=pb = p, and c=3c = 3.

Thus, we have:

  • α+β=p3\alpha + \beta = -\frac{p}{3} (Sum of the roots)
  • αβ=33=1\alpha \beta = \frac{3}{3} = 1 (Product of the roots)

Step 3: Substitute α=β2\alpha = \beta^2 into the equations.

From the sum of the roots: α+β=p3\alpha + \beta = -\frac{p}{3} Substitute α=β2\alpha = \beta^2: β2+β=p3\beta^2 + \beta = -\frac{p}{3} Thus, we have: β2+β+p3=0(Equation 1)\beta^2 + \beta + \frac{p}{3} = 0 \quad \text{(Equation 1)}

From the product of the roots: αβ=1\alpha \beta = 1 Substitute α=β2\alpha = \beta^2: β2β=1\beta^2 \cdot \beta = 1 This simplifies to: β3=1\beta^3 = 1 Thus, β=1\beta = 1 (since β=1\beta = 1 is the only real root of β3=1\beta^3 = 1).

Step 4: Solve for pp.

Now that we know β=1\beta = 1, substitute this value into Equation 1: 12+1+p3=01^2 + 1 + \frac{p}{3} = 0 1+1+p3=01 + 1 + \frac{p}{3} = 0 2+p3=02 + \frac{p}{3} = 0 p3=2\frac{p}{3} = -2 p=6p = -6

Step 5: Conclusion.

Since p>0p > 0 is given in the problem, it seems that there was an inconsistency in the problem statement, as p=6p = -6 is negative. If the condition p>0p > 0 is strict, we would need to revisit the problem.

Do you want to go through the steps in greater detail or clarify any part of this solution?

Follow-up Questions:

  1. How do you apply Vieta’s formulas to different types of quadratic equations?
  2. What would happen if the condition p>0p > 0 was adjusted to allow for p=6p = -6?
  3. Can you explore the behavior of cubic equations with real roots?
  4. How do you interpret the relationship between the sum and product of roots in a quadratic equation?
  5. What is the geometric interpretation of quadratic equations with roots that are related by squaring?

Tip:

For quadratic equations where roots are related, always consider substitution (like α=β2\alpha = \beta^2) and use Vieta's relations to set up solvable equations.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Vieta's Formulas

Formulas

Sum of roots: α + β = -b/a
Product of roots: αβ = c/a

Theorems

Vieta's Formulas

Suitable Grade Level

Grades 10-12