If the roots of the equation x^3-11x^2+36x-36 | vedclass

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(C) The given equation is $x^3-11x^2+36x-36=0$. Factoring the cubic equation,we get $(x-2)(x-3)(x-6)=0$. Thus,the ex-radii are $r_1=2, r_2=3, r_3=6$.

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) The given equation is $x^3-11x^2+36x-36=0$. Factoring the cubic equation,we get $(x-2)(x-3)(x-6)=0$. Thus,the ex-radii are $r_1=2, r_2=3, r_3=6$. We know that $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$. Since $r = \frac{\Delta}{s} = 1$,we have $\Delta = s$. Using $r_1 = \frac{\Delta}{s-a} = \frac{s}{s-a} = 2$,we get $s = 2s - 2a$,so $2a = s$. Using $r_2 = \frac{s}{s-b} = 3$,we get $s = 3s - 3b$,so $3b = 2s$. Using $r_3 = \frac{s}{s-c} = 6$,we get $s = 6s - 6c$,so $6c = 5s$. Since $s = \frac{a+b+c}{2}$,we have $2s = a+b+c$. Substituting $a = \frac{s}{2}, b = \frac{2s}{3}, c = \frac{5s}{6}$ into $a+b+c = 2s$: $\frac{s}{2} + \frac{2s}{3} + \frac{5s}{6} = \frac{3s+4s+5s}{6} = \frac{12s}{6} = 2s$. This holds for any $s$. To find the perimeter $2s$,we use $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = s$. $\sqrt{s(s-\frac{s}{2})(s-\frac{2s}{3})(s-\frac{5s}{6})} = s$ $\Rightarrow \sqrt{s(\frac{s}{2})(\frac{s}{3})(\frac{s}{6})} = s$ $\Rightarrow \sqrt{\frac{s^4}{36}} = s$ $\Rightarrow \frac{s^2}{6} = s$. Since $s 
eq 0$,$s = 6$. Therefore,the perimeter $2s = 2(6) = 12$.

If the roots of the equation $x^3-11x^2+36x-36=0$ are the ex-radii of a triangle $ABC$,then the perimeter of the triangle $ABC$ is

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