Quadratic Equations
Symmetric functions of roots
Grade 11
Question:
<p>If \(\alpha, \beta\) are the roots of the equation \(ax^2 + bx + c = 0\), then the value of \(\dfrac{(a\alpha^2 + c)}{(a\alpha + b)} + \dfrac{(a\beta^2 + c)}{(a\beta + b)}\) is</p>
<p>\(\dfrac{b(b^2 - 2ac)}{4a}\)</p>
<p>\(\dfrac{b^2 - 4ac}{2a}\)</p>
<p>\(\dfrac{b(b^2 - 2ac)}{a^2c}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Since α and β are roots of ax² + bx + c = 0, we have aα² + bα + c = 0, which gives aα² + c = -bα. Use this substitution to simplify each fraction drastically.
<p><strong>Step 1:</strong> Since α is a root of ax² + bx + c = 0, we have:</p><p>aα² + bα + c = 0 ⟹ aα² + c = -bα</p><p>Similarly, aβ² + c = -bβ</p><p><strong>Step 2:</strong> Substitute into the first fraction:</p><p>$$\frac{a\alpha^2 + c}{a\alpha + b} = \frac{-b\alpha}{a\alpha + b}$$</p><p>$$\frac{a\beta^2 + c}{a\beta + b} = \frac{-b\beta}{a\beta + b}$$</p><p><strong>Step 3:</strong> The sum becomes:</p><p>$$\frac{-b\alpha}{a\alpha + b} + \frac{-b\beta}{a\beta + b} = -b\left(\frac{\alpha}{a\alpha + b} + \frac{\beta}{a\beta + b}\right)$$</p><p><strong>Step 4:</strong> Combine fractions:</p><p>$$= -b\left(\frac{\alpha(a\beta + b) + \beta(a\alpha + b)}{(a\alpha + b)(a\beta + b)}\right)$$</p><p>$$= -b\left(\frac{a\alpha\beta + b\alpha + a\alpha\beta + b\beta}{(a\alpha + b)(a\beta + b)}\right)$$</p><p>$$= -b\left(\frac{2a\alpha\beta + b(\alpha + \beta)}{(a\alpha + b)(a\beta + b)}\right)$$</p><p><strong>Step 5:</strong> The denominator expands to: a²αβ + ab(α + β) + b²</p><p>By Vieta's formulas: α + β = -b/a and αβ = c/a</p><p>Numerator: 2ac + b(-b) = 2ac - b²</p><p>Denominator: a²(c/a) + ab(-b/a) + b² = ac - b² + b² = ac</p><p><strong>Step 6:</strong> Therefore:</p><p>$$= -b\cdot\frac{2ac - b^2}{ac} = \frac{-b(2ac - b^2)}{ac} = \frac{b^3 - 2abc}{ac}$$</p><p>∴ Answer: C</p>
Correct Answer: C