Question

Find the quadratic polynomial whose zeros are 3+√5/2 and 3-√5/2​

Answer 1

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]}

k[x

2

−(α+β)x+αβ]

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

5

),β=(3−

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

−(α+β)x+αβ]

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

2

−(6)x+4]

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

2

−6x+4

Therefore the quadratic polynomial is :-

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

2

−6x+4