Question:
Zeros of $p(x)=x^{2}-2 x-3$ are
(a) 1, −3
(b) 3, −1
(c) −3, −1
(d) 1, 3
Solution:
(b) 3,-1
Here, $\mathrm{p}(\mathrm{x})=x^{2}-2 x-3$
Let $x^{2}-2 x-3=0$
$=>x^{2}-(3-1) x-3=0$
$=>x^{2}-3 x+x-3=0$
$=>x(x-3)+1(x-3)=0$
$=>(x-3)(x+1)=0$
$=>x=3,-1$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.