Question

if the zeroes of the cubic polynomial f(x) = kx3-8x2=5 are alpha-beta, alpha and alpha+beta then find the value of k​

Answer 1

Answer:

f(x) = kx³ – 8x² + 5

Roots are α – β , α & α +β

Sum of roots = – (-8)/k

Sum of roots = α – β + α + α +β = 3α

= 3α = 8/k

= k = 8/3α

or we can solve as below

f(x) = (x – (α – β)(x – α)(x – (α +β))

= (x – α)(x² – x(α+β + α – β) + (α² – β²))

= (x – α)(x² – 2xα + (α² – β²))

= x³ – 2x²α + x(α² – β²) – αx² +2α²x – α³ + αβ²

= x³ – 3αx² + x(3α² – β²) + αβ² – α³

= kx³ – 3αkx² + xk(3α² – β²) + k(αβ² – α³)

comparing with

kx³ – 8x² + 5

k(3α² – β²) = 0 => 3α² = β²

k(αβ² – α³) = 5

=k(3α³ – α³) = 5

= k2α³ = 5

3αk = 8 => k = 8/3α

(8/3α)2α³ = 5

=> α² = 15/16

=> α = √15 / 4

Answer 2

$\huge\bold\blue{⠀⠀⠀♡Answer♡⠀⠀⠀}$

⠀⠀⠀⠀⠀☆F(x) =Kx³ – 8x² = 5☆

➠Roots are a – β, a & a + β

➠Sum of roots = –(-8)/k

➠Sum of roots = a – β + a + a + β = 3a

= 3α = 8/k

= 3α = 8/k= k = 8/3α

or we can solve as below

↬f(x) = (x – (α – β)(x – α)(x – (α +β))

↬ (x – α)(x² – x(α+β + α – β) + (α² – β²))

↬ (x – α)(x² – 2xα + (α² – β²))

↬ x³ – 2x²α + x(α² – β²) – αx² +2α²x – α³ + αβ²

↬ x³ – 3αx² + x(3α² – β²) + αβ² – α³

↬ kx³ – 3αkx² + xk(3α² – β²) + k(αβ² – α³)

comparing with

➵ kx³ – 8x² + 5

➵ k(3α² – β²) = 0 => 3α² = β²

➵ k(αβ² – α³) = 5

➵ k(3α³ – α³) = 5

➵ k2α³ = 5

➠3αk = 8 => k = 8/3α

➠(8/3α)2α³ = 5

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀➜ α² = 15/16

⠀⠀⠀⠀⠀⠀⠀➜ α = √15 / 4