• # question_answer Let, a be a non-zero real number and a, p be the roots of the equation $a{{x}^{2}}+5x+2=0$.Then, the absolute value of the difference of the roots of the equation${{a}^{3}}{{(x+5)}^{2}}-25a(x+5)+50=0$ A)  $|{{\alpha }^{2}}-{{\beta }^{2}}|$      B)  $|\alpha \beta \,({{\alpha }^{2}}-{{\beta }^{2}})|$ C) $\left| \frac{{{\alpha }^{2}}-{{\beta }^{2}}}{\alpha \beta } \right|$                 D)  $\left| \frac{{{\alpha }^{2}}-{{\beta }^{2}}}{{{\alpha }^{2}}{{\beta }^{2}}} \right|$

Since, $SD=\sqrt{{{60}^{2}}+{{25}^{2}}}$ and $=\sqrt{4225}=65=DP$ are the roots of $\Delta x=(SA+AP)-SP$ $\Rightarrow$ and $\Delta x=(65+65)-120$ are roots of the transformed equation $\Rightarrow$ Let $\Delta x=10\,m$ and $\frac{\lambda }{2}$ be roots of $\lambda$ $=\left( 10-\frac{\lambda }{2} \right)$ and $=(2n)\frac{\lambda }{2}$ all roots of $n=0,\,\,1,\,\,2,...$ Difference between roots $10-\frac{\lambda }{2}=(2n)\frac{\lambda }{2},\,\,n=0,\,\,1,\,2,...$ $10=(2n+1)\frac{\lambda }{2},\,n=0,\,1,\,\,2,...$ $\Rightarrow$c $\lambda =\frac{20}{2n+1},\,\,n=0,\,\,1,\,\,2,...$ $\Rightarrow$ $\lambda =20,\,\frac{20}{3},\,\frac{20}{5},\,\frac{20}{7},...$ $\phi$