Question

find a quadratic polynomial whose zeroes are 3+√5/5 and 3-√5/5​

Answer 1

Answer:

$25x { }^{2} - 30x + 4 = 0$

Step-by-step explanation:

consider the zeroes be X

now

the first zero

$x = \frac{3 + \sqrt{5} }{5}$

solving

$5x - 3 - \sqrt{5} = 0.....(1)$

now the second zero

$x = \frac{3 - \sqrt{5} }{5}$

if we solve it we get

$5x - 3 + \sqrt{5} = 0.....eq(2)$

multiplying equation (1) and (2)

$(5x - 3 - \sqrt{5} )(5x - 3 + \sqrt{5} ) = 0$

using identities

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

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

$(5x - 3) {}^{2} - ( \sqrt{5} ) {}^{2}$

$25x {}^{2} + 9 - 30x - 5 = 0$

$25x {}^{2} - 30x + 4 = 0$

Answer 2

Answer:

=x²-6x-116

Step-by-step explanation:

zeroes of polynomial are 3+√5/5 and 3-√5/5​

let α=3+5√5 , β=3-5√5

so α+β=3+5√5+3-5√5=6

αβ =(3+5√5)(3-5√5)

=3²-(5√5)²=9-125=-116

the polynomial with zeros α and β is:

f(x)=x²-(α+β)x+αβ

=x²-6x-116