Let a cubic equation x 3+ ax 2+ bx + c =0 has roots α, β, γ. If 2 is added to each, then the results are reciprocal of square of original roots, then

Question

Let a cubic equation x 3+ ax 2+ bx + c =0 has roots α, β, &gamma;. If 2 is added to each, then the results are reciprocal of square of original roots, then (A) a + b + c =1  (B) a +2 b + c =1 (C)  abc = a +2 b +2 c (D) abc =2 a +2 b + c

Tardigrade Admin · Accepted Answer

∵ α+2=(1/α2) ⇒ α3+2 α2-1=0 ∵ x 3+ ax 2+ bx + c =0 and x 3+2 x 2-1=0 have all roots in common. ∴ a =2, b =0, c =-1.

Q. Let a cubic equation $x^{3} + a x^{2} + b x + c = 0$ has roots $α, β, γ$ . If 2 is added to each, then the results are reciprocal of square of original roots, then

$a + b + c = 1$

$a + 2 b + c = 1$

$ab c = a + 2 b + 2 c$

$ab c = 2 a + 2 b + c$

$∵ α + 2 = \frac{1}{α ^{2}} \Rightarrow α^{3} + 2 α^{2} - 1 = 0$
$∵ x^{3} + a x^{2} + b x + c = 0$ and $x^{3} + 2 x^{2} - 1 = 0$ have all roots in common.
$∴ a = 2, b = 0, c = - 1$ .

Q. Let a cubic equation x3+ax2+bx+c=0 has roots α,β,γ. If 2 is added to each, then the results are reciprocal of square of original roots, then

a+b+c=1

a+2b+c=1

abc=a+2b+2c

abc=2a+2b+c