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Q.

If R is the circum radius and ${r, r}_{1} {, r}_{2} {, r}_{3}$ are inradius and ex-radii of the triangle then the roots of the equation $x^{3} - (4 R + r) x^{2} + s^{2} x - r s^{2} = 0$ are

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a

$\frac{r_{1}}{r_{2}}, \frac{r_{2}}{r_{3}}, \frac{r_{3}}{r_{1}}$

b

$r_{1} + r_{2}, r_{2} + r_{3}, r_{3} + r_{1}$

c

$r_{1}, r_{2}, r_{3}$

d

$r_{1} r_{2}, r_{2} r_{3}, r_{3} r_{1}$

answer is B.

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Detailed Solution

$α + β + γ = 4 R + r$

we know that $r_{1} + r_{2} + r_{3} = 4 R + r$

$α β γ = r s^{2}$

roots are $\Rightarrow α = r_{1}, β = r_{2}, γ = r_{3}$

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If R is the circum radius and r, r1, r2, r3 are inradius and ex-radii of the triangle then the roots of the equation x3−(4R+r)x2+s2x−rs2=0 are