Math Problem Statement

α and β are the zeros of the quadratic polynomial x² - 5x + 6, find the value of α⁴ + β² + α² + β⁴.

Solution

The given quadratic polynomial is:

x25x+6=0x^2 - 5x + 6 = 0

Step 1: Identify sum and product of roots

From the standard form ax2+bx+c=0ax^2 + bx + c = 0, we compare:

  • Sum of roots: α+β=ba=51=5\alpha + \beta = -\frac{b}{a} = \frac{5}{1} = 5
  • Product of roots: αβ=ca=61=6\alpha \beta = \frac{c}{a} = \frac{6}{1} = 6

Step 2: Compute α4+β4+α2β2\alpha^4 + \beta^4 + \alpha^2\beta^2

Using the identity:

α4+β4=(α2+β2)22α2β2\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2

First, calculate α2+β2\alpha^2 + \beta^2 using:

α2+β2=(α+β)22αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta

=522(6)=2512=13= 5^2 - 2(6) = 25 - 12 = 13

Next, calculate α2β2\alpha^2\beta^2 using:

α2β2=(αβ)2=62=36\alpha^2\beta^2 = (\alpha \beta)^2 = 6^2 = 36

Now, substitute these values:

α4+β4=(13)22(36)=16972=97\alpha^4 + \beta^4 = (13)^2 - 2(36) = 169 - 72 = 97

Thus, the required value is:

α4+β4+α2β2=97+36=133\alpha^4 + \beta^4 + \alpha^2\beta^2 = 97 + 36 = \mathbf{133}


Would you like further clarification or step-by-step breakdowns? Here are some related questions to explore:

  1. How do we derive the identities used in solving this problem?
  2. Can we generalize this for any quadratic equation?
  3. What happens if the quadratic polynomial has complex roots?
  4. How do we extend this method to higher-degree polynomials?
  5. Can we solve this problem using a different approach?

Tip: Always verify your algebraic identities before applying them in calculations!

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Roots of Polynomials
Power of Terms

Formulas

Sum of roots: α + β = -b/a
Product of roots: αβ = c/a
Identity for powers: α⁴ + β⁴ = (α² + β²)² - 2α²β²

Theorems

Vieta's Formulas
Quadratic Identity

Suitable Grade Level

Grades 9-11