Quadratic Equations
Roots transformation
Grade 11
Question:
<p>If <em>α</em>, <em>β</em> are the roots of <em>ax</em><sup>2</sup> + <em>c</em> = <em>bx</em>, then the equation (<em>a</em> + <em>cy</em>)<sup>2</sup> = <em>b</em><sup>2</sup><em>y</em> in <em>y</em> has the roots</p>
<p>\(\alpha\beta^{-1},\ \alpha^{-1}\beta\)</p>
<p>\(\alpha^2,\ \beta^{-2}\)</p>
<p>\(\alpha^{-1},\ \beta^{-1}\)</p>
<p>\(\alpha^2,\ \beta^2\)</p>
Step-by-Step Solution
Key Concept: If α, β are roots of ax² + c = bx, then α + β = b/a and αβ = c/a. The equation in y represents a transformation where y = 1/x, so the roots of the y-equation are reciprocals of the original roots.
<p><strong>Step 1:</strong> Rewrite the original equation in standard form: ax² - bx + c = 0</p><p>If α, β are roots, then by Vieta's formulas: α + β = b/a and αβ = c/a</p><p><strong>Step 2:</strong> The equation (a + cy)² = b²y can be rewritten as: a² + 2acy + c²y² = b²y</p><p>Rearranging: c²y² + (2ac - b²)y + a² = 0</p><p><strong>Step 3:</strong> Divide by c²: y² + [(2ac - b²)/c²]y + (a²/c²) = 0</p><p><strong>Step 4:</strong> Notice that if we substitute y = 1/x into the original equation ax² - bx + c = 0, we get:</p><p>a(1/y)² - b(1/y) + c = 0, which simplifies to: c²y² + (2ac - b²)y + a² = 0</p><p><strong>Step 5:</strong> Therefore, the roots of the y-equation are y₁ = 1/α and y₂ = 1/β</p><p><strong>∴ Answer: The roots are 1/α and 1/β (or reciprocals of the original roots)</strong></p>
Correct Answer: D