Find the sum and product of the zeroes of the following quadratic polynomials.
$x^{2} - (\sqrt{2} + 1) x + \sqrt{2}$

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Solution

The standard form of any polynomial is $(coefficient of x^{2}) x^{2} + (coefficient of x) x + constant term$ .
Here, coefficient of $x^{2}$ is $1$ , coefficient of $x$ is $- (\sqrt{2} + 1)$ , and constant term is $\sqrt{2}$ .
Step 1: To find the sum of zeroes.
Sum of zeroes = $\frac{- (coefficient of x)}{coefficient of x^{2}} = \frac{- [- (\sqrt{2} + 1)]}{1} = \sqrt{2} + 1$
Step 2: To find the product of zeroes.
Product of zeroes = $\frac{constant term}{coefficient of x^{2}} = \frac{\sqrt{2}}{1} = \sqrt{2}$
Hence, the sum and product of zeroes are $\sqrt{2} + 1$ and $\sqrt{2}$ respectively.

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