Question

Determine k and solve the equation 2x³ – 6x² + 3x + k = 0 if one of its roots is twice the sum of the other two roots

Answer 1

Given cubic equation

2x³ – 6x² + 3x + k = 0

Let the roots be α, β, γ

Given α = 2(β + γ)

β + γ = `alpha/2`  .......(1)

Sum of roots α(β + γ) = 3

From (1) `alpha + alpha/2` = 3

`(3alpha)/2` = 3

⇒ α = 2

Again αβ + βγ + γα = `3/2`

α = 2 

⇒ 2β + βγ + 2γ = `3/2`

From (1) `2(alpha/2) + betaγ = 3/2`

βγ = `3/2 - 2`

= `(3 - 4)/2`

= `(-1)/2`

βγ = `(-1)/2`  ........(2)

Product of roots α β γ = `- "k"/2`

α = 2

⇒ 2βγ = `- "k"/2`

βγ = `- "k"/4`  .......(3)

From (2) and (3)

`- "k"/4 = (- 1)/2`

k = `4/2` = 2

k = 2

Also α = 2

⇒ x – 2 is a factor

Other factor `2x^2 - 2x - 1` = 0

x = `(2 +- sqrt(4 + 8))/4`

= `(2 +- 2sqrt(3))/4`

= `(2(1 +- sqrt(3)))/4`

= `(1 +- sqrt(3))/2`

∴ The roots are 2, `(1 + sqrt(3))/2, (1 - sqrt(3))/2`