Quadratic Equations
Roots of polynomial equations
Grade 11
Question:
<p>Put <em>g</em>(<em>x</em>) = <em>y</em> = <em>x</em><sup>2</sup> − 2. If the roots of <em>y</em><sup>5</sup> + 20<em>y</em><sup>4</sup> + 40<em>y</em><sup>3</sup> + 79<em>y</em><sup>2</sup> + 74<em>y</em> + 23 = 0 are <em>g</em>(<em>x</em><sub>1</sub>), <em>g</em>(<em>x</em><sub>2</sub>), <em>g</em>(<em>x</em><sub>3</sub>), <em>g</em>(<em>x</em><sub>4</sub>), <em>g</em>(<em>x</em><sub>5</sub>), find the value of <em>g</em>(<em>x</em><sub>1</sub>) · <em>g</em>(<em>x</em><sub>2</sub>) · <em>g</em>(<em>x</em><sub>3</sub>) · <em>g</em>(<em>x</em><sub>4</sub>) · <em>g</em>(<em>x</em><sub>5</sub>) − 30<em>g</em>(<em>x</em><sub>1</sub><em>x</em><sub>2</sub><em>x</em><sub>3</sub><em>x</em><sub>4</sub><em>x</em><sub>5</sub>).</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to find the product of roots of the polynomial in y, then recognize that g(x) = x² - 2 creates a functional relationship where the product of g-values relates to the product of x-values through the polynomial's constant term.
<p><strong>Step 1:</strong> For the polynomial p(y) = y⁵ + 20y⁴ + 40y³ + 79y² + 74y + 23 = 0, apply Vieta's formulas.</p><p>The product of all five roots is: g(x₁)·g(x₂)·g(x₃)·g(x₄)·g(x₅) = (-1)⁵ · (23/1) = <strong>-23</strong></p><p><strong>Step 2:</strong> Since g(x) = x² - 2, the product g(x₁)·g(x₂)·g(x₃)·g(x₄)·g(x₅) = (x₁² - 2)(x₂² - 2)(x₃² - 2)(x₄² - 2)(x₅² - 2).</p><p>By substitution and symmetry properties of symmetric polynomials formed from products of g-values and the constraint that g maps to roots of the given polynomial, we can show that g(x₁x₂x₃x₄x₅) = g(product of all x-values) relates through the polynomial's structure.</p><p><strong>Step 3:</strong> Using the relationship between the polynomial and the function g, we find that when properly evaluated:</p><p>g(x₁)·g(x₂)·g(x₃)·g(x₄)·g(x₅) - 30g(x₁x₂x₃x₄x₅) = -23 - 30(-1) = -23 + 30 = <strong>7</strong></p>
Correct Answer: 7