Question

Draw the graph of $y=x^2+3x+2$ and use it to solve the equation $x^2+2x+4=0$.

Answer 1

First, let us form a table for $y=x^2+3x+2$

$x$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$x^2$	$16$	$9$	$4$	$1$	$0$	$1$	$4$	$9$
$3x$	$-12$	$-9$	$-6$	$-3$	$0$	$3$	$6$	$9$
$2$	$2$	$2$	$2$	$2$	$2$	$2$	$2$	$2$
$y$	$6$	$2$	$0$	$0$	$2$	$6$	$12$	$20$

Plot the points $(-4,6),(-3,2),(-2,0)),(1,6),(2,12)$ and $(3,20)$
Now, join the points by a smooth curve. The curve so obtained, is the graph of $y=x^2+3x+2$.
Now, $x^2+2x+4=0$
$\Rightarrow$ $x^2+3x+2-x+2=0$
$\Rightarrow$ $y=x-2$ $\because$ $y=x^2+3x+2$
Thus the roots of $x^2+2x+4=0$ are obtained from the points of intersection of
$y=x-2$ and $y=x^2+3x+2$
Now, form the table for the line $y=x-2$

$x$	$-2$	$0$	$1$	$2$
$y=x-2$	$-4$	$-2$	$-1$	$0$

The straight line $y=x-2$ does not intersect the curve $y=x^2+3x+2$. Thus, $x^2+2x+4=0$ has not real roots.