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

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(D) The given equation is $x^2-5x+6=0$. \ Solving for $x$: $(x-3)(x-2)=0$,which gives $x=3$ and $x=2$. \ Let the two sides be $a=3$ and $b=2$,and th

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$5+\sqrt{7}$

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(D) The given equation is $x^2-5x+6=0$. \ Solving for $x$: $(x-3)(x-2)=0$,which gives $x=3$ and $x=2$. \ Let the two sides be $a=3$ and $b=2$,and the included angle be $C=\frac{\pi}{3}$. \ 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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