If frac{k}{kx+3}+frac{3}{3x-k}=frac{12x+5}{(k | vedclass

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(A) Given the equation: $\frac{k}{kx+3}+\frac{3}{3x-k}=\frac{12x+5}{(kx+3)(3x-k)}$Multiplying both sides by $(kx+3)(3x-k)$,we get:$k(3x-k) + 3(kx+3) =

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Rational numbers

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(A) Given the equation: $\frac{k}{kx+3}+\frac{3}{3x-k}=\frac{12x+5}{(kx+3)(3x-k)}$Multiplying both sides by $(kx+3)(3x-k)$,we get:$k(3x-k) + 3(kx+3) = 12x+5$$3kx - k^2 + 3kx + 9 = 12x + 5$$6kx - k^2 + 9 = 12x + 5$Comparing the coefficients of $x$ on both sides:$6k = 12 \Rightarrow k = 2$Substituting $k=2$ into the quadratic equation $kx^2-7x+3=0$:$2x^2-7x+3=0$$2x^2-6x-x+3=0$$2x(x-3)-1(x-3)=0$$(2x-1)(x-3)=0$Thus,the roots are $x = \frac{1}{2}$ and $x = 3$.Since both $\frac{1}{2}$ and $3$ are rational numbers,the correct option is $A$.

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

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