Answer:
Quadratic polynomial \[={{x}^{2}}-\] (Sum of zeroes) x + Product of zeroes \[={{x}^{2}}-(0)x+\left( \frac{-3}{5} \right)={{x}^{2}}-\frac{3}{5}\] \[={{(x)}^{2}}-\left( \sqrt{\frac{3}{5}} \right)\] =\[\left( x-\sqrt{\frac{3}{5}} \right)\left( x+\sqrt{\frac{3}{5}} \right)\left[ \begin{align} & \text{By}\,\,\text{applying} \\ & \left( {{a}^{2}}-{{b}^{2}} \right)=(a+b)(a-b) \\ \end{align} \right]\] Zeroes are, \[x-\sqrt{\frac{3}{5}}=0\] or \[x+\sqrt{\frac{3}{5}}=0\] \[\Rightarrow x=\sqrt{\frac{3}{5}}\] or \[\Rightarrow x=-\sqrt{\frac{3}{5}}\] \[x=\sqrt{\frac{3}{5}\times \frac{5}{3}}\] or \[x=-\sqrt{\frac{3}{5}\times \frac{5}{3}}\] \[x=\frac{\sqrt{15}}{5}\] or \[x=\frac{-\sqrt{15}}{5}\]
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