Two sides of a triangle are given by the root | vedclass

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(D) The given equation is $x^2-5x+6=0$. Factoring the equation,we get $(x-3)(x-2)=0$,which gives the roots $x=3$ and $x=2$. These roots represent the

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$5+\sqrt{7}$

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(D) The given equation is $x^2-5x+6=0$. Factoring the equation,we get $(x-3)(x-2)=0$,which gives the roots $x=3$ and $x=2$. These roots represent the two sides of the triangle,so let $a=3$ and $b=2$. The angle between these sides is $C = \frac{\pi}{3}$. Using the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$. Substituting the values,$\cos(\frac{\pi}{3}) = \frac{3^2+2^2-c^2}{2 	imes 3 	imes 2}$. $\frac{1}{2} = \frac{9+4-c^2}{12} \Rightarrow \frac{1}{2} = \frac{13-c^2}{12}$. $6 = 13-c^2$ $\Rightarrow c^2 = 7$ $\Rightarrow c = \sqrt{7}$. The perimeter of the triangle is $a+b+c = 3+2+\sqrt{7} = 5+\sqrt{7}$.

Two sides of a triangle are given by the roots of the equation $x^2-5x+6=0$ and the angle between the sides is $\frac{\pi}{3}$. Then,the perimeter of the triangle is

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