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Find the zeroes of the quadratic polynomial $\left(x^{2}-5\right)$ and verify the relation between the zeroes and the coefficients.
Find the zeroes of the quadratic polynomial $\left(x^{2}-5\right)$ and verify the relation between the zeroes and the coefficients.
CBSE | Class 10 | Excercise 2A | Maths | Polynomials | RS Aggarwal
Find the zeroes of the quadratic polynomial $\left(x^{2}-5\right)$ and verify the relation between the zeroes and the coefficients.

$$
\begin{aligned}
&f(x)=x^{2}-5 \\
&\text { It can be written as } x^{2}+0 x-5 . \\
&=\left(x^{2}-(\sqrt{5})^{2}\right) \\
&=(x+\sqrt{5})(x-\sqrt{5}) \\
&\therefore f(x)=0 \Rightarrow(x+\sqrt{5})(x-\sqrt{5})=0 \\
&\Rightarrow x+\sqrt{5}=0 \text { or } x-\sqrt{5}=0 \\
&\Rightarrow x=-\sqrt{5} \text { or } x=\sqrt{5}
\end{aligned}
$$

=> the zeroes of $f(x)$ are $-\sqrt{5}$ and $\sqrt{5}$.

Since, the coefficient of $\mathrm{x}$ is 0 and the coefficient of $x^{2}$ is 1

Sum of zeroes $=-\sqrt{5}+\sqrt{5}=\frac{0}{1}=\frac{-(\text { coefficient of } x)}{\left(\text { coefficient of } x^{2}\right)}$

Product of zeroes $=-\sqrt{5} \times \sqrt{5}=\frac{-5}{1}=\frac{\text { constant term }}{\left(\text { coef ficient of } x^{2}\right)}$

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