Quadratic Equations
Roots in HP
Grade 11

Question:

<p>Given that \(\alpha, \beta\) are roots of the equation \(Ax^2 - 4x + 1 = 0\) and \(\gamma, \delta\) are the roots of the equation \(Bx^2 - 6x + 1 = 0\), such that \(\alpha, \beta, \gamma\) and \(\delta\) are in HP then</p>
<p>(a) \(A = 3\)</p>
<p>(b) \(A = 4\)</p>
<p>(c) \(B = 2\)</p>
<p>(d) \(B = 8\)</p>

Step-by-Step Solution

Key Concept: If numbers are in HP, their reciprocals are in AP. Transform the given equations by substituting $x = 1/y$ to get equations with reciprocal roots.
<p><strong>Solution:</strong> Since $\alpha, \beta, \gamma$ and $\delta$ are in HP, hence $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ and $\frac{1}{\delta}$ are in AP and they may be taken as $a-3d, a-d, a+d$ and $a+3d$.</p><p>Replacing $x$ by $\frac{1}{x}$, we get the equation whose roots are $a-3d, a-d, a+d, a+3d$ is $x^2 - 4x + A = 0$.</p><p>∴ The correct options are (a) and (d).</p>
Correct Answer: a,d

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