Question

25 Find a quadratic polynomial whose zeroes are 5-3 V2 and 5 +3√2​

Answer 1

Answer:

x²-10x+7

Step-by-step explanation:

α=5-3√2 , β=5+3√2

x= 5-3√2 , x= 5+3√2

[x-(5-3√2)]=0 , [x-(5+3√2)]=0

(x-5+3√2)=0 and (x-5-3√2)=0 are the factors of the polynomial

Hence the quadratic equation= (x-5+3√2)(x-5-3√2)

                                                 = (x-5)² - (3√2)²= x²-10x + 25 - 18

                                                 = x²-10x+7

Answer 2

Let:

$\alpha = 5 - 3 \sqrt{2} \: and \: \beta = 5 + \sqrt{3}$

Now,

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

$= > \alpha \beta = ({5})^{2} </strong><strong>-</strong><strong> {( \sqrt{3} )}^{2}$

[tex] = > \alpha \beta = 25 - 3

[tex] = > \alpha \beta = 22

Also,

[tex] \alpha + \beta = 5 + \sqrt{3} + 5 - \sqrt{3} [/tex]

[tex] = > \alpha + \beta = 10[/tex]

Now, we know that:

Quadratic polynomial = k {x² - (sum of zeroes)x + (product of zeroes)}

[tex] = k({ {x}^{2} - 10x + 22})

where, k is a constant.

Hence, this is the required answer.