<p>If <em>x</em><sup>2</sup> + <em>px</em> + <em>q</em> = 0 and <em>x</em><sup>2</sup> + <em>qx</em> + <em>p</em> = 0, (<em>p</em> ≠ <em>q</em>) have a common root, show that 1 + <em>p</em> + <em>q</em> = 0. Also, show that their other roots are the roots of the equation <em>x</em><sup>2</sup> + <em>x</em> + <em>pq</em> = 0.</p>
Step-by-Step Solution
Key Concept: If two quadratic equations share a common root, that root must satisfy both equations simultaneously. Subtracting the equations eliminates x², giving a linear relationship that constrains p and q. The common root must also satisfy the resulting equation, which forces 1 + p + q = 0.
<p><strong>Step 1:</strong> Let α be the common root of both equations.</p><p>Then: α² + pα + q = 0 ... (1) and α² + qα + p = 0 ... (2)</p><p><strong>Step 2:</strong> Subtract equation (2) from equation (1):</p><p>α² + pα + q - (α² + qα + p) = 0</p><p>α(p - q) + (q - p) = 0</p><p>α(p - q) - (p - q) = 0</p><p>(p - q)(α - 1) = 0</p><p><strong>Step 3:</strong> Since p ≠ q, we have (p - q) ≠ 0, therefore <strong>α = 1</strong></p><p><strong>Step 4:</strong> Substitute α = 1 in equation (1):</p><p>1 + p + q = 0 ... ✓ <strong>Proved</strong></p><p><strong>Step 5:</strong> For the first equation x² + px + q = 0 with roots α = 1 and β:</p><p>By Vieta's formulas: 1 + β = -p and 1·β = q</p><p>Therefore: β = -p - 1 = q (using 1 + p + q = 0)</p><p><strong>Step 6:</strong> For the second equation x² + qx + p = 0 with roots α = 1 and γ:</p><p>By Vieta's formulas: 1 + γ = -q and 1·γ = p</p><p>Therefore: γ = -q - 1 = p (using 1 + p + q = 0)</p><p><strong>Step 7:</strong> The other roots are β = q and γ = p. These satisfy:</p><p>Sum = q + p, Product = q·p = pq</p><p>The equation with these roots: <strong>x² - (p + q)x + pq = 0</strong></p><p>Since 1 + p + q = 0, we have p + q = -1</p><p>∴ <strong>x² + x + pq = 0</strong> ... ✓ <strong>Proved</strong></p>
Correct Answer: 1 + p + q = 0