Question

Find a quadratic polynomial whose zeroes are (3+√5) and (3-√5) ​

Answer 1

GIVEN :

Zeroes of Quadratic Polynomial = (3+√5) and (3-√5)

Let α = (3+√5) β = (3-√5)

We know that,

Quadratic Polynomial = x² - (sum of zeroes)x + (product of zeroes)

Sum of zeroes :

=> α + β

=> (3+√5) + (3-√5)

=> 3 + √5 + 3 - √5

=> 6

α + β = 6

Product of zeroes :

=> αβ

=> (3+√5) (3-√5)

=> 3² - (√5)²

=> 9 - 5

=> 4

αβ = 4

Therefore, the Quadratic Polynomial is :

=> x² - (α + β)x + αβ

=> x² - 6x + 4

Answer 2

Answer:-

Given,

(3+√5) and (3-√5) are the zeros of the polynomial.

To find the quadratic polynomial we use :-

$\boxed{k[ {x}^{2} - ( \alpha + \beta )x + \alpha \beta]}$

$Let \: \alpha = (3 + \sqrt{5}) , \beta = (3 - \sqrt{5)}$

Sum of the zeros = α+β

= (3+√5)+(3-√5)

= 3+√5+3-√5

= 6

Therefore sum of the zeros (α+β) = 6

Product of the zeros = αβ

= (3+√5)(3-√5)

= 4

Therefore product of the zeros (αβ) = 4

Now substitute the values to find the zeros of the polynomial.

${k[ {x}^{2} - ( \alpha + \beta )x + \alpha \beta]}$

${=k[ {x}^{2} - (6)x + 4]}$

${={x}^{2} - 6x + 4}$

Therefore the quadratic polynomial is :-

${{x}^{2} - 6x + 4}$