Question

Find the co-ordinates points where the graph of polynomial $x^2+x+12$ intersects the x-axis.

Answer 1

Let ,

$y=x^2+x+12=0$

It cut x-axis then, at x-axis $y=0$

$x^2+x+12=0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-\left(1\right)\pm \sqrt{(1{)}^2-4\ast 1\ast 12}}{2\ast 1}$

$x=\frac{-1\pm \sqrt{-47}}{2}$

Hence required coordinates.

$(x,y)=(\frac{-1+\sqrt{-47}}{2},0)$

$(x,y)=(\frac{-1-\sqrt{-47}}{2},0)$