Quadratic Equations
Roots and Coefficients
Grade 11
Question:
<p>If \(\cos^4 A + p, \sin^4 A + p\) are the roots of the equation \(x^2 + a(2x + 1) = 0\) and \(\cos^2 A + q, \sin^2 A + q\) are the roots of the equation \(x^2 + 4x + 2 = 0\) then \(a\) is equal to</p>
<p>(a) \(-2\)</p>
<p>(b) \(-1\)</p>
<p>(c) \(1\)</p>
<p>(d) \(2\)</p>
Step-by-Step Solution
Key Concept: Use algebraic identities for sum of fourth and second powers, and Vieta's formulas to relate the roots and coefficients of both equations.
<p><strong>Solution:</strong> Since $\cos^4 A + \sin^4 A = \cos^2 A + \sin^2 A - 2\cos^2 A \sin^2 A = 1 - 2\cos^2 A \sin^2 A$</p><p>And $(\cos^4 A + p) + (\sin^4 A + p) = (\cos^2 A + q) + (\sin^2 A + q)$</p><p>From Vieta's formulas: sum of roots of first equation = $-2a$ and sum of roots of second equation = $-4$</p><p>For the second equation: $\cos^2 A + \sin^2 A + 2q = 1 + 2q = -4$, so $q = -\frac{5}{2}$</p><p>For the first equation: $\cos^4 A + \sin^4 A + 2p = -2a$</p><p>This gives: $4a^2 - 4a = 8$ or $a^2 - a - 2 = 0$</p><p>Therefore $(a-2)(a+1) = 0$ or $a = 2, -1$</p><p>∴ The correct options are (b) and (d).</p>
Correct Answer: b,d