Question

Find the zero of the quadratic polynomial $x^2-3$, and verify the relationship between the zeros and coefficients.

Answer 1

Let $p\left(x\right)=x^2-3$

Zero of the polynomial is the value of $x$ where $p\left(x\right)=0$

Put $p\left(x\right)=0\Rightarrow x^2-3=0$

$\Rightarrow x^2-{\left(\sqrt{3}\right)}^2=0$

So,$x=\sqrt{3},-\sqrt{3}$

$\therefore \alpha =\sqrt{3}$ and $\beta =-\sqrt{3}$ are zeroes of the polynomial.

We can write $p\left(x\right)=x^2-3=x^2+0-3$ is of the form $a{x}^2+bx+c$ where $a=1,b=0,c=-3$

L.H.S$=$Sum of the zeroes$=\alpha +\beta =\sqrt{3}-\sqrt{3}=0$

and R.H.S$=$Sum of the zeroes$=\frac{-b}{a}=\frac{-0}{1}=0$

L.H.S$=$Product of the zeroes$=\alpha \beta =\sqrt{3}\times -\sqrt{3}=-3$

and R.H.S$=$product of the zeroes$=\frac{c}{a}=\frac{-3}{1}=-3$

Since L.H.S$=$R.H.S

Hence relationship between zeroes and coefficient is verified.