Quadratic Equations
Nature of Roots
Grade 11
Question:
<p>For Problems 28–30: The numbers \(a\), \(b\), and \(c\) are between 2 and 18, such that (i) their sum is 25, (ii) the numbers 2, \(a\), and \(b\) are consecutive terms of an A.P., (iii) the numbers \(b\), \(c\), 18 are consecutive terms of a G.P.<br>Roots of the equation \(ax^2 + bx + c = 0\) are</p>
<p>(1) real and positive</p>
<p>(2) real and negative</p>
<p>(3) imaginary</p>
<p>(4) real and of opposite sign</p>
Step-by-Step Solution
Key Concept: Use the A.P. and G.P. conditions to express a and b in terms of a common difference, then apply the sum constraint to find exact values before analyzing the quadratic's discriminant.
<p><strong>Step 1: Use A.P. condition</strong><br>Since 2, a, b are in A.P.: a − 2 = b − a<br>Therefore: <strong>b = 2a − 2</strong></p><p><strong>Step 2: Use G.P. condition</strong><br>Since b, c, 18 are in G.P.: c/b = 18/c<br>Therefore: <strong>c² = 18b</strong></p><p><strong>Step 3: Apply sum constraint</strong><br>a + b + c = 25<br>Substitute b = 2a − 2:<br>a + (2a − 2) + c = 25<br>3a + c = 27, so <strong>c = 27 − 3a</strong></p><p><strong>Step 4: Solve for a</strong><br>From c² = 18b and c = 27 − 3a, b = 2a − 2:<br>(27 − 3a)² = 18(2a − 2)<br>729 − 162a + 9a² = 36a − 36<br>9a² − 198a + 765 = 0<br>a² − 22a + 85 = 0<br>(a − 5)(a − 17) = 0<br>So a = 5 or a = 17</p><p><strong>Step 5: Check validity</strong><br>If a = 5: b = 8, c = 12 ✓ (all in [2,18])<br>If a = 17: b = 32 ✗ (exceeds 18)<br>Therefore: <strong>a = 5, b = 8, c = 12</strong></p><p><strong>Step 6: Analyze roots of 5x² + 8x + 12 = 0</strong><br>Discriminant: Δ = 64 − 4(5)(12) = 64 − 240 = −176 < 0<br>Since Δ < 0, the roots are <strong>complex conjugates</strong><br>∴ Answer: <strong>Complex conjugate roots</strong> (or <strong>3</strong> if option codes complex/imaginary as option 3)</p>
Correct Answer: 3