Question

find the quadaratic polynomial if the zeroes are 2+√3/3 abd 2-√3/3

Answer 1

let the zeroes be x

$x = \frac{2 + \sqrt{3} }{3}$

$x = \frac{2 - \sqrt{3} }{3}$

$3x - 2 - \sqrt{3 } = 0......(1)$

$3x - 2 + \sqrt{3 } = 0......(2)$

multipling (1) and (2)

$(3x - 2 - \sqrt{3} )(3x - 2 + \sqrt{3} ) = 0$

using

$(a + b)(a - b) = a { }^{2} - b {}^{2}$

$(3x - 2) {}^{2} - ( \sqrt{3} ) { = 0}$

$9x {}^{2} + 4 - 12x - 3 = 0$

$9x {}^{2} - 12x + 1 = 0$

.

$\large{vergilneeestobeDie}$

Answer 2

Answer:

The polynomial is k(9x² - 12x + 1) , where 'k' is a constant.

Given:

• zeroes are 2+√3/3 and 2-√3/3

To find:

• The polynomial with this zeros.

Solving Question:

If and α are β the zeros then,

α + β = -b/a

α * β = c/a

We are given the zeros of the polynomial,therefore we can substitute the values in it to find the answer.

Solution:

α = (2 + √3)/3

β = (2-√3)/3

⇒α + β = -b/a

or, (2+√3)/3 +(2-√3)/3 = -b/a

or, 4/3 = -b/a ........equ(1)

[Multiply 3 so as to make the denominators of equ(1) and equ(2) same ]

⇒ 4/3 * 3/3 = 12/9 = -b/a

⇒ a = 9 ; b = -12

α * β = c/a

or, [(2-√3)/3]* [(2+√3)/3]

or, 2² -(√3)²/9   [∵ (a+b)(a-b) = a² - b² ]

or, 4 -3 /9

or, 1/9  = c/a .........equ(2)

⇒ c = 1

standard form = ax² +bx +c =0

∴ The polynomial is k(9x² - 12x + 1) , where 'k' is a constant.