Quadratic Equations
Nature and Properties of Roots
Grade 11
Question:
<p>Let \(p, q \in \mathbb{R}\). If \(2 - \sqrt{3}\) is a root of the quadratic equation \(x^2 + px + q = 0\), then</p>
<p>(a) \(q^2 - 4p - 16 = 0\)</p>
<p>(b) \(p^2 - 4q - 12 = 0\)</p>
<p>(c) \(p^2 - 4q + 12 = 0\)</p>
<p>(d) \(q^2 + 4p + 14 = 0\)</p>
Step-by-Step Solution
Key Concept: When a quadratic equation with real coefficients has an irrational root of the form $a + \sqrt{b}$, its conjugate $a - \sqrt{b}$ is also a root. Use Vieta's formulas to find the parameters.
<p><strong>Solution:</strong></p><p>Given quadratic equation is $x^2 + px + q = 0$ with $p, q \in \mathbb{R}$ having one root $2 - \sqrt{3}$.</p><p>Since coefficients are real and one root is $2 - \sqrt{3}$, the other root is $2 + \sqrt{3}$ (conjugate of $2 - \sqrt{3}$).</p><p>Sum of roots: $-p = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$</p><p>Therefore, $p = -4$</p><p>Product of roots: $q = (2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$</p><p>Therefore, $q = 1$</p><p>Now checking option (b): $p^2 - 4q - 12 = (-4)^2 - 4(1) - 12 = 16 - 4 - 12 = 0$ ✓</p><p>∴ Answer is (b).</p>
Correct Answer: B