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Question

Find the condition that x3px2+qxr=0 may here
(1) two roots equal but of opposite sign;
(2) the roots in geometrical progression.

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Solution

Given, x3px2+qxr=0

Case 1: Roots are α,α,β

Sum of the roots, S1=αα+β=pβ=p

Sum of the roots taken 2 at a time, S2=α2+αβαβ=qα2=q

Product of the roots, $S_3 = -\alpha^2\beta = r

From the above equations we can see that S1S3=S2pq=r

Case 2: Roots are in geometric progression. Let the roots be α,β,γ

αβ=βγ

β2=αγβ3=αβγ=r

Since, β is the roots of the given equation, we have

β3pβ2+qβr=0rpr2/3+qr1/3r=0

p3r=q3[taking cubes on both sides and simplifying]


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