Question

If $\alpha \ne \beta$ but ${\alpha }^2=5\alpha -3,{\beta }^2=5\beta -3$ then find the equation whose roots are $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$.

Answer 1

${\alpha }^2=5\alpha -3;{\beta }^2=5\beta -3$

${\alpha }^2-5\alpha +3=0;{\beta }^2-5\beta +3=0$

$\alpha =\frac{5\pm \sqrt{25-12}}{2}\beta =\frac{5\pm \sqrt{25-12}}{2}$

$\therefore \alpha ,\beta =\frac{5\pm \sqrt{13}}{2}$

$\therefore$ equation as $\frac{\alpha }{\beta },\frac{\beta }{\alpha }$

$\therefore$ product of roots $=\frac{\alpha }{\beta },\frac{\beta }{\alpha }=1$

sum of roots $=\frac{\alpha }{\beta }=1$

$=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{(\alpha +\beta {)}^2}{\alpha \beta }2\alpha \beta$

$\frac{{\left(\frac{5\pm \sqrt{3}+5-\sqrt{3}}{2}\right)}^2-2\times \frac{1}{4}(25-13)}{\frac{1}{4}(25-13)}$

$=\frac{(5{)}^2-\frac{1}{2}\times 12}{\frac{1}{4}\times 12}[\because \alpha .\beta =\frac{1}{4}(5+\sqrt{13})(5-\sqrt{13})=\frac{1}{2}(5^2-{\left(\sqrt{13}\right)}^2)]$

$=\frac{25-6}{3}=\frac{19}{3}$

equation is : $x^2-\frac{19}{3}x+1=0$

or $3x^2-19x+3=0$