Question

If alpha and beta are the zeroes of polynomial such that alpha +beta =6 and alpha*beta = 4 then write the quadratic polynomial

Answer 1

We know that to form a quadratic equation the form is
x2-Sx+P,
where s=sum
p=product
so equation is x2-6x=4 
 

Answer 2

Solutions 

Alpha = α
Beta = β

We recall that for a quadratic equation in the form 

y = ax² + bx + c 

The sum and product of the roots can be determined as such: 

y = x^2 - (sum)x + (product) 

The numerical coefficient of x² must be 1. 

Applying this on the given function, 

5y² - 7y + 1 


(1/5) (y² - (7/5)y + 1/5) 

Therefore, sum of the roots (α + β) and product of the roots (αβ), respectively, are: 

α + β = 7/5 

αβ = 1/5 

Now, we are asked to find a polynomial with roots 2α/β and 2β/α. What we'll do is get their sum and product: 

Product: 

2α/β × 2β/α 

4αβ / αβ 

= 4 

Sum: 

2α/β + 2β/α 

(2α² + 2β²) / αβ

2(α² + β²) / αβ 

Complete the square inside the parenthesis by adding 2ab, but at the same time subtracting 2(2ab) outside the parenthesis to counter its effect: 

[ 2(α² + 2αβ + b^2) - 4αβ ] / αβ]
[ 2(α + β)² - 4αβ ] / αβ]

note that, we have values for α + β and αβ from above. 

[ 2(7/5)² - 4(1/5) ] / (1/5)]

= 78/5 

Therefore, the polynomial is, 

y² - (sum)y + (product) 

y² - (78/5)y + 4