Quadratic Equations
Roots and Coefficients
Grade 11
Question:
<p>If <i>a</i> and <i>b</i> are the roots of the equation <i>7x</i><sup>2</sup> – <i>3x</i> – <i>2</i> = <i>0</i>, then the value of \(\frac{a}{1-a^2} + \frac{b}{1-b^2}\) is equal to</p>
<p>(a) \(\frac{27}{32}\)</p>
<p>(b) \(\frac{1}{24}\)</p>
<p>(c) \(\frac{3}{8}\)</p>
<p>(d) \(\frac{27}{16}\)</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to find sum and product of roots, then simplify the algebraic expression by finding a common denominator.
<p><strong>Solution:</strong> Given quadratic equation <i>7x</i><sup>2</sup> – <i>3x</i> – <i>2</i> = <i>0</i> has roots <i>a</i> and <i>b</i>.</p><p>By Vieta's formulas: $a + b = \frac{3}{7}$ and $ab = -\frac{2}{7}$</p><p>$\frac{a}{1-a^2} + \frac{b}{1-b^2} = \frac{a(1-b^2) + b(1-a^2)}{(1-a^2)(1-b^2)}$</p><p>$= \frac{a - ab^2 + b - a^2b}{1 - (a^2 + b^2) + a^2b^2}$</p><p>$= \frac{(a+b) - ab(a+b)}{1 - [(a+b)^2 - 2ab] + (ab)^2}$</p><p>Numerator: $\frac{3}{7} - (-\frac{2}{7}) \cdot \frac{3}{7} = \frac{3}{7} + \frac{6}{49} = \frac{21 + 6}{49} = \frac{27}{49}$</p><p>Denominator: $1 - [(\frac{3}{7})^2 + \frac{4}{7}] + \frac{4}{49} = 1 - [\frac{9}{49} + \frac{4}{7}] + \frac{4}{49} = 1 - \frac{9+28}{49} + \frac{4}{49} = \frac{49 - 37 + 4}{49} = \frac{16}{49}$</p><p>$\therefore \frac{27/49}{16/49} = \frac{27}{16}$</p>
Correct Answer: D