Question

If α and β are the zeros of the quadratic polynomial p(s) = 3s² - 6s + 4 , find the value of (α/β) + (β/α) + 2(1/α + 1/β) + 3αβ ,

Answer 1

(α/β) + (β/α) + 2(1/α + 1/β) + 3αβ = 8

Step-by-step explanation:

p(s) = 3s² - 6s + 4

α and β are the zeroes of the polynominial.

Then α + β = -b/a = -(-6)/3 = 2

αβ = c/a  = 4/3

Now  (α/β) + (β/α) + 2(1/α + 1/β) + 3αβ  

= (α² + β²)/ αβ + 2 (α + β/ αβ) + 3αβ  

= [ (α + β)² - 2αβ + 2(α + β) ] / αβ + 3αβ  

= [ 2² - 2(4/3) + 2(2) ] / (4/3) + 3(4/3)

= (4 -8/3 + 4)/(4/3) + 4

= 16/3 / 4/3 + 4

= 16/4 + 4

= 4 + 4

= 8

Value of (α/β) + (β/α) + 2(1/α + 1/β) + 3αβ = 8