Quadratic Equations
Polynomial equations and root properties
Grade 11
Question:
<p>Let \(\alpha, \beta, \gamma, \delta\) be roots of \(x^4 - 12x^3 + lx^2 - 54x + 14 = 0\). If \(\alpha + \beta = \gamma + \delta\), then</p>
<p>(a) \(l = 45\)</p>
<p>(b) \(l = -45\)</p>
<p>(c) If \(\alpha^2 + \beta^2 < \gamma^2 + \delta^2\) then \(\frac{ab}{gd} = \frac{7}{2}\)</p>
<p>(d) If \(\alpha^2 + \beta^2 < \gamma^2 + \delta^2\) then \(\frac{ab}{gd} = \frac{2}{7}\)</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas with the constraint that two pairs of roots sum to equal values.
<p>By Vieta's formulas: $\alpha + \beta + \gamma + \delta = 12$. If $\alpha + \beta = \gamma + \delta$, then $\alpha + \beta = \gamma + \delta = 6$. From the sum of products of roots taken two at a time: $\alpha\beta + \gamma\delta + (\alpha + \beta)(\gamma + \delta) = l$, so $\alpha\beta + \gamma\delta + 36 = l$. Product of all roots: $\alpha\beta\gamma\delta = 14$. Testing $l = 45$: $\alpha\beta + \gamma\delta = 9$. Using constraints and solving, if $\alpha^2 + \beta^2 < \gamma^2 + \delta^2$, then $\frac{\alpha\beta}{\gamma\delta} = \frac{7}{2}$.</p>
Correct Answer: A, C