Question

if tye zeroes of cubic polynomial f(x) kx³-8x²+5 are a-b, a , a+b, find the value of k. correct answer will be marked as BRAINLIEST ANSWER​

Answer 1

Answer:

Answer:

k = 8/3α

Step-by-step explanation:

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

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