Form a quadratic polynomial whose zeroes are $3+\sqrt{2}$ and $ 3-\sqrt{2}$.

Given: Two zeroes $3+\sqrt{2}$ and $ 3-\sqrt{2}$.

To do: To form a quadratic polynomial with the given zeroes.

Solution:

 Here, sum of the zeroes

 $S=( \alpha +\beta ) $

 $=\left( 3+\sqrt{2}\right) +\left( 3-\sqrt{2}\right) =6$

Product of the given zeroes,

$P=( \alpha \times \beta )$

$=\left( 3+\sqrt{2}\right) \times \left( 3-\sqrt{2}\right) $

$=3^{2} -\left(\sqrt{2}\right)^{2} =9-2=7$

 As known, If a $\alpha $ and $\beta$ are two zeroes of a quadratic polynomial, 

then the polynomial is,$x^{2} -Sx+P$

On subtituting the value of $S$ and $P$, we have

$x^{2} -6x+7$

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Updated on: 10-Oct-2022

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