Step 1: Find the zeroes.
Solve $x^2-5=0 $\Rightarrow$ x=\pm\sqrt{5}$. Thus the zeroes are $\alpha=\sqrt{5}$ and $\beta=-\sqrt{5}$.
Step 2: Verify relationships.
For $ax^2+bx+c$, sum of zeroes $=\,-\dfrac{b}{a}$ and product $=\,\dfrac{c}{a}$.
Here $a=1,\ b=0,\ c=-5$.
Sum: $\alpha+\beta=\sqrt{5}+(-\sqrt{5})=0\ = -\dfrac{b}{a}=-\dfrac{0}{1}=0$.
Product: $\alpha\beta=\sqrt{5}\cdot(-\sqrt{5})=-5\ = \dfrac{c}{a}=\dfrac{-5}{1}=-5$.
\boxed{\text{Zeroes: }\sqrt{5},\ -\sqrt{5}\ \ \text{and both relations are verified.}}
The product of $\sqrt{2}$ and $(2-\sqrt{2})$ will be:
If a tangent $PQ$ at a point $P$ of a circle of radius $5 \,\text{cm}$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12 \,\text{cm}$, then length of $PQ$ will be:
In the figure $DE \parallel BC$. If $AD = 3\,\text{cm}$, $DE = 4\,\text{cm}$ and $DB = 1.5\,\text{cm}$, then the measure of $BC$ will be: