The zeroes of the polynomial $4 \mathrm{x}^{2}+5 \sqrt{2} \mathrm{x}-3$ are (a) $-3 \sqrt{2}, \sqrt{2}$ (b) $-3 \sqrt{2}, \frac{\sqrt{2}}{2}$ (c) $\frac{-3}{\sqrt{2}}, \frac{\sqrt{2}}{4}$ (d) none of these

The correct option is option (c) $\frac{3}{\sqrt{2}}, \frac{\sqrt{2}}{4}$

$f(x)=4 x^{2}+5 \sqrt{2} x-3=0$

$\Rightarrow 4 x^{2}+6 \sqrt{2} x-\sqrt{2} x-3=0$

$\Rightarrow 2 \sqrt{2} \mathrm{x}(\sqrt{2} \mathrm{x}+3)-1(\sqrt{2} \mathrm{x}+3)=0$

$\Rightarrow(\sqrt{2} x+3)(2 \sqrt{2} x-1)=0$

$\Rightarrow \mathrm{x}=-\frac{3}{\sqrt{2}}$ or $\mathrm{x}=\frac{1}{2 \sqrt{2}}$

$\Rightarrow \mathrm{x}=-\frac{3}{\sqrt{2}}$ or $\mathrm{x}=\frac{1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{4}$