• # question_answer Case Study : Q. 26 to 30 Ramesh was asked by one of his friends to find the polynomial whose zeroes are $\frac{-2}{\sqrt{3}}$ and $\frac{\sqrt{3}}{4}$. He obtained the polynomial as shown below: Let $\alpha =\frac{-2}{\sqrt{3}}$ and $\alpha =\frac{\sqrt{3}}{4}$ $\Rightarrow \,\,\,\,\,\,\,\alpha +\beta =\frac{-2}{\sqrt{3}}+\frac{\sqrt{3}}{4}=\frac{-8+1}{4\sqrt{3}}=\frac{-7}{4\sqrt{3}}$ $\Rightarrow \,\,\,\,\,\,\,\alpha \beta =\frac{-2}{\sqrt{3}}\times \frac{\sqrt{3}}{4}=\frac{-1}{2}$ Required polynomial $={{x}^{2}}-(\alpha +\beta )x+\alpha \beta$ $={{x}^{2}}-\left( \frac{-7}{4\sqrt{3}} \right)x+\left( \frac{-1}{2} \right)$ $={{x}^{2}}+\frac{7x}{4\sqrt{3}}-\frac{1}{2}$ $=4\sqrt{3}{{x}^{2}}+7x-2\sqrt{3}$ His another friend Kavita pointed out that the polynomial obtained is not correct. Based on the above situation, answer the following questions: Is the claim of Kavita correct? A) Yes B) No C) Can't say D) None of these

 Given, $\alpha =\frac{-2}{\sqrt{3}}$and $\beta =\frac{\sqrt{3}}{4}$ $\alpha +\beta =\frac{-2}{\sqrt{3}}+\frac{\sqrt{3}}{4}=\frac{-8+3}{4\sqrt{3}}=\frac{-5}{4\sqrt{3}}$ $\alpha \beta =\frac{-2}{\sqrt{3}}\times \frac{\sqrt{3}}{4}=-\frac{1}{2}$ Yes, because value of $(\alpha +\beta )$ calculated by Anirudh is incorrect. So, option [a] is correct.