The cubic equation whose roots are thrice to each of the roots of x3 + 2x2 - 4x + 1 = 0 is

Question

The cubic equation whose roots are thrice to each of the roots of x3 + 2x2 - 4x + 1 = 0 is (A) x3 - 6x2 + 36x + 27 = 0 (B) x3 + 6x2 + 36x + 27 = 0 (C) x3 - 6x2 - 36x + 27 = 0 (D) x3 + 6x2 - 36x + 27 = 0

Tardigrade Admin · Accepted Answer

Given equation is x3+2 x2-4 x+1=0 Let α, β and &gamma; be the roots of the given equation ∴ α+β+&gamma;=-2, α β+β &gamma;+&gamma; α=-4 and α β &gamma;=-1 Let the required cubic equation has the roots 3 α, 3 β and 3 &gamma;. ⇒ 3 α+3 β+3 &gamma;=-6, 3 α .3 β+3 β+3 &gamma;+3 &gamma; .3 α=36 and 3 α .3 β .3 &gamma;=-27 ∴ Required equation is x3-(-6) x2+(-36) x-(-27)=0 ⇒ x3+6 x2-36 x+27=0

Q. The cubic equation whose roots are thrice to each of the roots of $x^{3} + 2 x^{2} - 4 x + 1 = 0$ is

$x^{3} - 6 x^{2} + 36 x + 27 = 0$

$x^{3} + 6 x^{2} + 36 x + 27 = 0$

$x^{3} - 6 x^{2} - 36 x + 27 = 0$

$x^{3} + 6 x^{2} - 36 x + 27 = 0$

Q. The cubic equation whose roots are thrice to each of the roots of x3+2x2−4x+1=0 is

x3−6x2+36x+27=0

x3+6x2+36x+27=0

x3−6x2−36x+27=0

x3+6x2−36x+27=0

Q. The cubic equation whose roots are thrice to each of the roots of $x^{3} + 2 x^{2} - 4 x + 1 = 0$ is

$x^{3} - 6 x^{2} + 36 x + 27 = 0$

$x^{3} + 6 x^{2} + 36 x + 27 = 0$

$x^{3} - 6 x^{2} - 36 x + 27 = 0$

$x^{3} + 6 x^{2} - 36 x + 27 = 0$