Question

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

• A. {1,2}
• B. {5,-6}
• C. cannot be determined
• D. No real roots

Answer 1

The correct option is D No real roots

First, form a table for $y=x^2+3x+2$.
$\begin{array}{ccccccccc}x& -4& -3& -2& -1& 0& 1& 2& 3\\ x^2& 16& 9& 4& 1& 0& 1& 4& 9\\ 3x& -12& -9& -6& -3& 0& 3& 6& 9\\ 2& 2& 2& 2& 2& 2& 2& 2& 2\\ y& 6& 2& 0& 0& 2& 6& 12& 20\end{array}$
Plot the points (-4, 6), (-3, 2), (-2, 0), (0, 2), (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 form the points of intersection of $y=x-2$ and $y=x^2+3x+2$.
Let us draw the graph of the straight line $y=x-2$.
Now, form the table for the line $y=x-2$
$\begin{array}{ccccc}x& -2& 0& 1& 2\\ y=x-2& -4& -2& -1& 0\end{array}$
The straight line $y=x-2$ does not intersect the curve $y=x^2+3x+2$.
Thus, $x^2+2x+4=0$ has no real roots.