Math Problem Statement

If alpha and beta are the zeroes of the quadratic polynomial, find a quadratic polynomial whose zeroes are 2alpha/beta and 2beta/alpha

Solution

Let the quadratic polynomial be represented as:

p(x)=ax2+bx+cp(x) = ax^2 + bx + c

Given that α\alpha and β\beta are the zeroes of this polynomial, we know the following relationships from Vieta's formulas:

  1. The sum of the roots α+β=ba\alpha + \beta = -\frac{b}{a}
  2. The product of the roots αβ=ca\alpha \beta = \frac{c}{a}

We are asked to find a new quadratic polynomial whose zeroes are 2αβ\frac{2\alpha}{\beta} and 2βα\frac{2\beta}{\alpha}.

To do this, we need to compute:

  • The sum of the new zeroes:
    Sum of new roots=2αβ+2βα=2(αβ+βα)\text{Sum of new roots} = \frac{2\alpha}{\beta} + \frac{2\beta}{\alpha} = 2\left( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \right) Using the identity αβ+βα=α2+β2αβ\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta}, we can express the sum as: Sum of new roots=2(α2+β2αβ)\text{Sum of new roots} = 2\left( \frac{\alpha^2 + \beta^2}{\alpha \beta} \right) Now, recall that: α2+β2=(α+β)22αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta Substitute α+β=ba\alpha + \beta = -\frac{b}{a} and αβ=ca\alpha \beta = \frac{c}{a} into the equation: α2+β2=(ba)22(ca)\alpha^2 + \beta^2 = \left( -\frac{b}{a} \right)^2 - 2\left( \frac{c}{a} \right) α2+β2=b2a22ca\alpha^2 + \beta^2 = \frac{b^2}{a^2} - \frac{2c}{a} Thus, the sum of the new roots becomes: Sum of new roots=2b2a22caca=2b22acac\text{Sum of new roots} = 2 \cdot \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c}{a}} = 2 \cdot \frac{b^2 - 2ac}{ac}

  • The product of the new zeroes:
    Product of new roots=2αβ2βα=4αβαβ=4\text{Product of new roots} = \frac{2\alpha}{\beta} \cdot \frac{2\beta}{\alpha} = \frac{4\alpha\beta}{\alpha\beta} = 4

Thus, the new quadratic polynomial has the form:

p(x)=x2(Sum of new roots)x+(Product of new roots)p(x) = x^2 - (\text{Sum of new roots}) \cdot x + (\text{Product of new roots})

Substituting the values for the sum and product:

p(x)=x2(2b22acac)x+4p(x) = x^2 - \left( 2 \cdot \frac{b^2 - 2ac}{ac} \right) x + 4

This is the required quadratic polynomial.


Would you like further details or have any questions on the steps?

Here are five related questions you might find helpful:

  1. What are Vieta's formulas and how do they relate to the zeroes of a polynomial?
  2. How do you simplify algebraic expressions involving fractions like αβ+βα\frac{\alpha}{\beta} + \frac{\beta}{\alpha}?
  3. Can you derive the relationship between the sum of the squares of the roots and the sum and product of the roots in a quadratic polynomial?
  4. How can you verify that the quadratic polynomial you derived has the correct zeroes?
  5. What would happen to the polynomial if the coefficients aa, bb, and cc were given specific values?

Tip: To simplify problems involving fractions with variables, always look for common denominators or factorization strategies!

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Vieta's Formulas
Roots of Polynomials

Formulas

Vieta's formulas: Sum of roots = -b/a, Product of roots = c/a
Sum of new roots = 2 * (alpha/beta + beta/alpha)
alpha^2 + beta^2 = (alpha + beta)^2 - 2*alpha*beta

Theorems

Vieta's Formulas
Sum and product of roots of quadratic equations

Suitable Grade Level

Grades 10-12