Quadratic Equations
Formation of Quadratic Equations
Grade 11
Question:
<p>Given \(p, q, r\) are real numbers (\(p \neq q,\ r \neq 0\)) and
\[\frac{1}{x+p} + \frac{1}{x+q} = \frac{1}{r}\]
The roots \(\alpha\) and \(\beta\) of the resulting quadratic satisfy which general form?</p>
<p>\(x^2 + (\alpha+\beta)x + \alpha\beta = 0\)</p>
<p>\(x^2 + (p+q-2r)x + pq - pr - qr = 0\)</p>
<p>\(x^2 - (p+q)x + pq = 0\)</p>
<p>\(x^2 + (p+q+2r)x + pq + pr + qr = 0\)</p>
Step-by-Step Solution
Key Concept: Clear the fractions by multiplying through by r(x+p)(x+q), then rearrange into standard quadratic form to identify the relationship between coefficients and roots via Vieta's formulas.
<p><strong>Step 1:</strong> Start with the given equation and find a common denominator on the left side:</p><p>$$\frac{1}{x+p} + \frac{1}{x+q} = \frac{1}{r}$$</p><p>$$\frac{(x+q) + (x+p)}{(x+p)(x+q)} = \frac{1}{r}$$</p><p>$$\frac{2x + p + q}{(x+p)(x+q)} = \frac{1}{r}$$</p><p><strong>Step 2:</strong> Cross-multiply to eliminate fractions:</p><p>$$r(2x + p + q) = (x+p)(x+q)$$</p><p><strong>Step 3:</strong> Expand the right side and rearrange into standard form:</p><p>$$r(2x + p + q) = x^2 + (p+q)x + pq$$</p><p>$$2rx + r(p+q) = x^2 + (p+q)x + pq$$</p><p>$$x^2 + (p+q-2r)x + (pq - r(p+q)) = 0$$</p><p><strong>Step 4:</strong> By Vieta's formulas for roots α and β:</p><p>$$\alpha + \beta = -(p+q-2r) = 2r - (p+q)$$</p><p>$$\alpha\beta = pq - r(p+q)$$</p><p>The roots satisfy: <strong>$$x^2 + (p+q-2r)x + (pq - r(p+q)) = 0$$</strong></p><p>∴ Answer: B</p>
Correct Answer: B