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Question

Solve the equations:
27x4195x3+494x2520x+192=0, the roots being in geometrical progression.

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Solution

Given equation, 27x4195x3+494x2520x+192=0 and let the roots be a,b,c,d

The roots are in Geometric Progression, therefore

ba=cb=dcad=bc

We know that abcd=19227ad=bc=83

(xa)(xd)=x2(a+d)x+ad=x2Ax+83, where A=a+d

(xb)(xc)=x2(b+c)x+bc=x2Bx+83, where B=b+c

x4659x3+49427x252027x+649=(xa)(xb)(xc)(xd)

=(x2Ax+83)(x2Bx+83).............(1)

=x4(A+B)x3+(AB+163)x2(83)(A+B)x+649

Comparing the coefficients of x and x2, we have

A+B=5202738=659 and AB=49427163=35027

Eliminating B in the above equations, we have

27A2+195A+350=0

(3A+10)(9A+35)=0A=359andB=103

Substituting value of A and B in (1), we have

(x2+359x+83)(x2+103x+83)=0

(9x2+35x+24)(3x2+10x+8)=0

(x+3)(9x+8)(3x+4)(x+2)=0

roots of the given equation are: 3,2,43,89


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