Q.If the sum of the squares of the roots of the equation \(6x^2 + x + k = 0\) is \(\frac{25}{36}\), then the value of \(k\) will be \(12\).

The statement is **false**.
For the equation \(6x^2 + x + k = 0\), let the roots be \(\alpha, \beta\). Then: - \(\alpha + \beta = -\frac{1}{6}\) - \(\alpha \beta = \frac{k}{6}\) According to the question: \[ \alpha^2 + \beta^2 = \frac{25}{36} \] Now using the identity: \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \] Substitute: \[ \left(-\frac{1}{6}\right)^2 - 2 \cdot \frac{k}{6} = \frac{25}{36} \Rightarrow \frac{1}{36} - \frac{k}{3} = \frac{25}{36} \Rightarrow -\frac{k}{3} = \frac{24}{36} = \frac{2}{3} \Rightarrow k = -2 \]
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