Question

find the quadritic polynomial whose zeroes are(3+√5)and(3-√5)​

Answer 1

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✤ Required Answer:

✒ GiveN:

• Zeroes of the polynomial are (3 + √5) and (3 - √5)

✒ To FinD:

• The required polynomial...$?$

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✤ How to find?

For this question, we need to know the relation between zeroes of the polynomial and the polynomial.

• Let α and β be the zeroes of a polynomial, then the polynomial is in the form:

✳ f(x) = x² - (α + β)x + αβ

Or, we can say that, a polynomial is denoted by:

✳ x² - (sum of its zeroes)x + product of its zeroes.

So, now we can find our Required polynomial with the help of the above relation.

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✤ Solution:

We have,

• Zeroes = (3 + √5) and (3 - √5)

According to above relation,

❒ Polynomial = x² - (sum of zeroes)x + product of zeroes

➙ x² - [(3 + √5) + (3 - √5)]x + (3 + √5)(3 - √5)

➙ x² - (3 + √5 + 3 - √5)x + 3² - (√5)²

➙ x² - 6x + 9 - 5

➙ x² - 6x + 4

✒ Required polynomial = x² - 6x + 4.

❍ Hence, solved!

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Answer 2

♣️ QUESTIONS ♣️

find the quadritic polynomial whose zeroes are(3+√5)and(3-√5).

♣️ANSWER♣️

Given:

zeroes of quadratic polynomial is (3+√5) and (3-√5).

Solution:

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

$sum \: of \: zeroes \: = \alpha + \beta = 3 + \sqrt{5} + 3 - \sqrt{5} = 6$

$\: product \: of \: zeroes = \alpha \beta = (3 + \sqrt{5} )(3 - \sqrt{5} ) = 9 - 5 = 4$

$sum \: of \: roots \: = \frac{ - b}{a} = \frac{ - 6}{1}$

$product \: of \: roots \: = \frac{c}{a} = \frac{4}{1}$

$\implies \: a = 1 \: and \: b = - 6 \: and \: c = 4$

Hence the quadratic polynomial is x²-6x+4.