Question

find the quadratic polynomials whose zeroes are 3+√2 and 3-√2

Answer 1

WE HAVE,

$\alpha =3+ \sqrt{2}$

$\beta =3- \sqrt{2}$

Now we have

$\alpha + \beta = 3+ \sqrt{2} + 3 - \sqrt{2} = 6$

$\alpha \beta = (3+ \sqrt{2} )(3- \sqrt{2} )$
  
                           = $9-3 \sqrt{2}+3 \sqrt{2}-2$

                           = $7$

Therefore polynomial = $x^{2} -6x+7$

PLZ MARK AS BRAINLIEST

Answer 2

Answer:

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Answer :

The required quadratic polynomial is :

x² - 6x + 7

Given :

The zeroes of a quadratic polynomial are : 3 + √2 and 3 - √2

To Find :

The quadratic polynomial having the given zeroes

Formula to be used :

If sum and product of zeroes of a quadratic polynomial are given then the polynomial can be expressed as :

$\rm x^{2} -(Sum \: of \: the \: zeroes)x + Product \: of \: the \: zeroes$

Solution :

Given the zeroes 3 + √2 and 3 - √2

$\rm \: Sum \: of \: the \: zeroes = 3 + \sqrt{2} + 3 - \sqrt{2} \\\\ \rm \implies Sum \: of \: the \: zeroes = 6$

Again ,

$\rm Product \: of \: the \: zeroes = (3+\sqrt{2})(3-\sqrt{2}) \\\\ \rm \implies Product \: of \: the \: zeroes = 3^{2} -(\sqrt{2})^{2} \\\\ \rm \implies Product \: of \: the \: zeroes = 9 - 2 \\\\ \rm \implies Product \: of \: the \: zeroes= 7$

Thus the quadratic equation is :

$\rm \dashrightarrow x^{2} - 6x + 7$