Quadratic Equations
Roots transformation
Grade None
Question:
<p>If <em>α</em> and <em>β</em> are roots of the equation <em>ax</em><sup>2</sup> + <em>bx</em> + <em>c</em> = 0, then the roots of the equation \(a(2x+1)^2 - b(2x+1)(3-x) + c(3-x)^2 = 0\) are</p>
<p>\(\dfrac{2\alpha+1}{\alpha-3},\ \dfrac{2\beta+1}{\beta-3}\)</p>
<p>\(\dfrac{3\alpha+1}{\alpha-2},\ \dfrac{3\beta+1}{\beta-2}\)</p>
<p>\(\dfrac{2\alpha-1}{\alpha-2},\ \dfrac{2\beta+1}{\beta-2}\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Recognize that the new equation has the form a·u² - b·u·v + c·v² = 0 where u = 2x+1 and v = 3-x. This is a homogeneous equation in u and v, so dividing by v² gives a quadratic in u/v that has the same roots as the original equation.
<p><strong>Step 1:</strong> Let u = 2x+1 and v = 3-x. The equation becomes au² - buv + cv² = 0.</p><p><strong>Step 2:</strong> Divide throughout by v²: a(u/v)² - b(u/v) + c = 0.</p><p><strong>Step 3:</strong> Let t = u/v. Then at² - bt + c = 0. Since α and β satisfy ax² - bx + c = 0 (which is equivalent since the coefficient of x² is a), we have t = α or t = β.</p><p><strong>Step 4:</strong> If (2x+1)/(3-x) = α, then 2x+1 = α(3-x), so 2x + αx = 3α - 1, giving x = (3α-1)/(2+α).</p><p><strong>Step 5:</strong> If (2x+1)/(3-x) = β, then similarly x = (3β-1)/(2+β).</p><p><strong>Step 6:</strong> Therefore, the roots are <strong>x = (3α-1)/(α+2) and x = (3β-1)/(β+2)</strong>, which correspond to option A.</p>
Correct Answer: A