If α , β, γ are the roots of the equation 2x3-3x2+6x+1=0 then α2 +β2+γ2 is equal to

Question

If α , β, &gamma; are the roots of the equation 2x3-3x2+6x+1=0 then α2 +β2+&gamma;2 is equal to (A) -(15/4) (B) (15/4) (C) (9/4) (D) 4

Tardigrade Admin · Accepted Answer

Since, α, β, &gamma; are the roots of the equation 2 x3-3 x2+6 x+1=0, then α+β+&gamma; =+(3/2) dots(i) α β+β &gamma;+&gamma; α =3 dots(ii) α β &gamma; =-(1/2) dots(iii) On squaring both sides Eq. (i), we get α2+β2+&gamma;2+2(α β+β &gamma;+&gamma; α)=(9/4) α2+β2+&gamma;2 =(9/4)-2(3) [from Eq. (ii) ] =(9/4)-6=-(15/4)

Q. If $α, β, γ$ are the roots of the equation $2 x^{3} - 3 x^{2} + 6 x + 1 = 0$ then $α^{2} + β^{2} + γ^{2}$ is equal to

$- \frac{15}{4}$

$\frac{15}{4}$

$\frac{9}{4}$

$4$

Q. If α,β,γ are the roots of the equation 2x3−3x2+6x+1=0 then α2+β2+γ2 is equal to

−415​

415​

49​

4

Since, α,β,γ are the roots of the equation 2x3−3x2+6x+1=0, then α+β+γ=+23​…(i) αβ+βγ+γα=3…(ii) αβγ=−21​…(iii) On squaring both sides Eq. (i), we get α2+β2+γ2+2(αβ+βγ+γα)=49​ α2+β2+γ2=49​−2(3) [from Eq. (ii) ] =49​−6=−415​

Q. If $α, β, γ$ are the roots of the equation $2 x^{3} - 3 x^{2} + 6 x + 1 = 0$ then $α^{2} + β^{2} + γ^{2}$ is equal to

$- \frac{15}{4}$

$\frac{15}{4}$

$\frac{9}{4}$

$4$