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(B) Given equation is $\frac{1}{2} x^{2}-\sqrt{11} x+1=0$.On comparing with $ax^{2}+bx+c=0$,we get:$a=\frac{1}{2}, b=-\sqrt{11}, c=1$.The quadratic fo

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$3+\sqrt{11}, \sqrt{11}-3$

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(B) Given equation is $\frac{1}{2} x^{2}-\sqrt{11} x+1=0$.On comparing with $ax^{2}+bx+c=0$,we get:$a=\frac{1}{2}, b=-\sqrt{11}, c=1$.The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.Substituting the values:$x = \frac{-(-\sqrt{11}) \pm \sqrt{(-\sqrt{11})^{2}-4 	imes \frac{1}{2} 	imes 1}}{2 	imes \frac{1}{2}}$$x = \frac{\sqrt{11} \pm \sqrt{11-2}}{1}$$x = \sqrt{11} \pm \sqrt{9}$$x = \sqrt{11} \pm 3$.Therefore,the roots are $3+\sqrt{11}$ and $\sqrt{11}-3$.

Find the roots of the quadratic equation by using the quadratic formula:
$\frac{1}{2} x^{2}-\sqrt{11} x+1=0$

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Find the roots of the quadratic equation by using the quadratic formula:$\frac{1}{2} x^{2}-\sqrt{11} x+1=0$