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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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