Question

# Solve the equations:27x4−195x3+494x2−520x+192=0, the roots being in geometrical progression.

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Solution

## Given equation, 27x4−195x3+494x2−520x+192=0 and let the roots be a,b,c,d The roots are in Geometric Progression, therefore ba=cb=dc⟹ad=bc We know that abcd=19227⟹ad=bc=83 (x−a)(x−d)=x2−(a+d)x+ad=x2–Ax+83, where A=a+d (x−b)(x−c)=x2−(b+c)x+bc=x2–Bx+83, where B=b+c x4−659x3+49427x2−52027x+649=(x−a)(x−b)(x−c)(x−d) =(x2–Ax+83)(x2–Bx+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=52027⋅38=−659 and AB=−49427−163=35027 Eliminating B in the above equations, we have 27A2+195A+350=0 ⟹(3A+10)(9A+35)=0⟹A=−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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