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

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x2-Sx+P,

where s=sum

p=product

so equation is x2-6x=4

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