Question

Find the Quaderatic polynomial whose roots are
3+√5 and 3-√5​

Answer 1

Answer:

$\: \: \tt{K( {x}^{2} -6x+4.)}$

Explanation:

$\: \: \: \: \: \: \tt{Let \: \alpha \: and \: \beta = 3 \pm \sqrt{5} .}$

So,

Sum of, it's Zeros.

$\tt{ \: \: \: \alpha + \beta = 3 + \cancel{\sqrt{5}} + 3 - \cancel{\sqrt{5}} .}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt { = 6.}$

Product of , it's Zeros.

$\: \: \: \tt{ \alpha \times \beta = 3 + \sqrt{5} \times 3 - \sqrt{5} . }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \: = 4.$

Now, Putting value Sum and Product be S , P respectively.

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{K( {x}^{2} - sx + p. )}$

$\leadsto \tt{K( {x}^{2} - 6x + 4. )}$

Let's Verify,

$\tt \: We \: have \: equation,$

$\: \: \: \tt{x}^{2} - 6x + 4.$

$\: \: \: \tt{d = {b}^{2} - 4ac.}$

$\: \: \: \: \: \: \tt = { - 6}^{2} - 4 \times 4 \times 1.$

$\: \: \: \: \: \: \tt= 36 - 16.$

$\: \: \: \: \: \: \tt = 20.$

$\tt x = \frac{6 \pm \sqrt{20} }{2}$

$\: \: \: \tt= \frac{6 \pm 2\sqrt{5} }{2}$

$\: \: \: \tt = \frac{ \cancel2(3 \pm \sqrt{5} )}{ \cancel2}$

$\tt x= 3 \pm \sqrt{5} .$

Therefore, The value both zeros be,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large\tt \alpha = 3 + \sqrt{5} .\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large\tt\beta = 3 - \sqrt{5} .$

Answer 2

AnswEr:

our Quadratic Polynomial is x² - 6x + 4.

ExplanaTion:

Given zeroes :

• 3 + √5
• 3 - √5

To find :

• Quadratic polynomial

Solution :

Sum of zeroes = $\sf{3\:+\:{\cancel{\sqrt{5}}}\:+\:3\:-\:{\cancel{\sqrt{5}}}}$

: $\implies$ Sum of zeroes = 3 + 3

: $\implies$ Sum of zeroes = 6

$\therefore$ Sum of given zeroes is 6.

$\rule{100}1$

Product of zeroes = ( 3 + √5 ) ( 3 - √5 )

We know that,

$\large{\boxed{\red{\sf{(a+b)(a-b)\:=\:a^2\:-\:b^2}}}}$

: $\implies$ Product of zeroes = (3)² - (√5)²

: $\implies$ 9 - 5

: $\implies$ 4

$\rule{200}2$

We also know that,

Quadratic polynomial = x² - Sx + P

Where, S refers to sum of zeroes and P refers to product of zeroes.

: $\implies$ Quadratic Polynomial = x² - 6x + 4

Hence, our Quadratic Polynomial is x² - 6x + 4.