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

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r1r2, r2r3, r3r1

r1, r2, r3

r1r2, r2r3, r3r1

r1+r2,r2+r3,r3+r1

answer is B.

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Detailed Solution

α+β+γ=4R+r we know that r1+r2+r3 =4R +rαβγ=rs2 roots are ⇒α=r1,β=r2,γ=r3

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