Question

Find the zeros of the following quadratic polynomial and verify the basic relationships between the zeros and the coefficients.
$x^2-2x-8$

• A. $(4,-2)$
• B. $(4,2)$
• C. $(-4,-2)$
• D. None of these

Answer 1

The correct option is A $(4,-2)$

$\Rightarrow$ The given quadratic equation is $x^2-2x-8=0$, comparing it with $a{x}^2+bx+c=0$

$\therefore$ We get, $a=1,b=-2,c=-8$

$x^2-2x-8=0$

$\Rightarrow$ $x^2-4x+2x-8=0$

$\Rightarrow$ $x(x-4)+2(x-4)=0$

$\Rightarrow$ $(x-4)(x+2)=0$

$\therefore$ $x=4$ and $x=-2$

$\Rightarrow$ $\alpha =4$ and $\beta =-2$

Now, we are going to verify relationship between zeros and the coefficient.

Sum of the zeros $=-\frac{Coefficientofx}{Coefficientof{x}^2}$

$\Rightarrow$ $\alpha +\beta =\frac{-b}{a}$

$\Rightarrow$ $4+(-2)=\frac{-(-2)}{1}$

$\Rightarrow$ $2=2$

$\therefore$ $L.H.S=R.H.S$

Product of the zeros $=\frac{Constantterm}{Coefficientof{x}^2}$

$\Rightarrow$ $\alpha .\beta =\frac{c}{a}$

$\Rightarrow$ $\left(4\right).(-2)=\frac{-8}{1}$

$\Rightarrow$ $-8=-8$

$\therefore$ $L.H.S=R.H.S$