Question

find a quadratic polynomial whose zeroes are 3+root 2and 3-root 2​

Answer 1

Solution

Given,

Let the zeros of the required quadratic polynomial be m and n

• m = 3 + √2
• n = 3 - √2

To finD

A quadratic polynomial

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Sum of Zeros : m + n

$\sf \: m \: + n = (3 + \sqrt{2} ) + (3 - \sqrt{2} ) \\ \\ \leadsto \: \sf \: m + n = 6 - - - - - - - (1)$

Product of Zeros : mn

$\sf \: mn = (3 + \sqrt{2} )(3 - \sqrt{2} ) \\ \\ \leadsto \: \sf \: mn = (3) {}^{2} - { (\sqrt{2} )}^{2} \\ \\ \leadsto \: \sf \: mn \: = 7 - - - - - - - (2)$

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Required Polynomial would be of the form :

$\sf \: p(x) = k \bigg( x {}^{2} - (m + n)x + mn \bigg) \: \\ \\ \implies \: \sf \: p(x) = k \: \bigg({x}^{2} - (6)x + 7 \bigg) \\ \\ \implies \: \boxed{ \boxed{\sf \: p(x) = {x}^{2} - 6x + 7}} \: \: \: \: \: \sf \: [lf \: k = 1]$

$\rule{300}{1}$

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Answer 2

Given: The zeros are 3 + √2 and 3 - √2.

To find: A polynomial that has 3 + √2 and 3 - √2 as its zeros.

Answer:

$\bf General\ form\ of\ a\ quadratic\ polynomial:\\\\\\\tt x^2\ -\ (Sum\ of\ the\ zeros)x\ +\ (Product\ of\ the\ zeros)$

From what we have,

$\sf Sum\ of\ the\ zeros:\\\\\\\implies\ \tt 3\ +\ \sqrt{2}\ +\ 3\ -\ \sqrt{2}\ =\ 3\ +\ 3\ =\ 6\\\\\\\sf Product\ of\ the\ zeros:\\\\\\\implies\ (3\ +\ \sqrt{2})\ \times\ (3 \ -\ \sqrt{2})\ =\ (3)^2\ -\ (\sqrt{2})^2\ =\ 9\ -\ 2\ =\ 7$

Using them in the general form,

$\tt x^2\ -\ 6x\ +\ 7$ is the required polynomial.