If $\frac{\mathrm{k}}{\mathrm{kx}+3}+\frac{3}{3 \mathrm{x}-\mathrm{k}}=\frac{12 \mathrm{x}+5}{(\mathrm{kx}+3)(3 \mathrm{x}-\mathrm{k})} \forall \mathrm{x} \in \mathrm{R}$ $-\left\{\frac{\{3\}}{\mathrm{k}}, \frac{\mathrm{k}}{3}\right\}$, then both the roots of the equation $\mathrm{kx}^2-7…

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If $\frac{\mathrm{k}}{\mathrm{kx}+3}+\frac{3}{3 \mathrm{x}-\mathrm{k}}=\frac{12 \mathrm{x}+5}{(\mathrm{kx}+3)(3 \mathrm{x}-\mathrm{k})} \forall \mathrm{x} \in \mathrm{R}$ $-\left\{\frac{\{3\}}{\mathrm{k}}, \frac{\mathrm{k}}{3}\right\}$, then both the roots of the equation $\mathrm{kx}^2-7 \mathrm{x}+3=0$ are

Aicharya · Accepted Answer

Given, $ \begin{aligned} & \frac{k}{(k x+3)}+\frac{3}{(3 x-k)}=\frac{12 x+5}{(k x+3)(3 x-k)} \ & \Rightarrow k(3 x-k)+3(k x+3)=12 x+5 \ & \Rightarrow 6 x k-k^2+9=12 x+5 \end{aligned} $ Comparing both side, we get $ \Rightarrow 6 k=12 \Rightarrow k=2 $ Hence, $k x^2-7 x+3=0$ $ \begin{aligned} \Rightarrow & 2 x^2-7 x+3=0 \ \Rightarrow \quad & 2 x^2-6 x-x+3=0 \ \Rightarrow \quad & 2 x(x-3)-1(x-3)=0 \ & (2 x-1)(x-3)=0 \ \Rightarrow \quad & x=\frac{1}{2}, 3 \end{aligned} $ Hence roots are rational numbers

If $\frac{\mathrm{k}}{\mathrm{kx}+3}+\frac{3}{3 \mathrm{x}-\mathrm{k}}=\frac{12…

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