Answer:
The equation contains exact roots at x = -4 and x = -1.
See attached image for the graph.
Step-by-step explanation:
We start by noticing that the expression on the left of the equal sign is a quadratic with leading term
, which means that its graph shows branches going up. Therefore:
1) if its vertex is ON the x axis, there would be one solution (root) to the equation.
2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.
3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.
So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:
We recall that the x-position of the vertex for a quadratic function of the form
is given by the expression: ![x_v=\frac{-b}{2a}](https://tex.z-dn.net/?f=x_v%3D%5Cfrac%7B-b%7D%7B2a%7D)
Since in our case
and
, we get that the x-position of the vertex is: ![x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}](https://tex.z-dn.net/?f=x_v%3D%5Cfrac%7B-b%7D%7B2a%7D%20%5C%5Cx_v%3D%5Cfrac%7B-5%7D%7B2%281%29%7D%5C%5Cx_v%3D-%5Cfrac%7B5%7D%7B2%7D)
Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:
![y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}](https://tex.z-dn.net/?f=y_v%3Df%28-%5Cfrac%7B5%7D%7B2%7D%29%5C%5Cy_v%3D%28-%5Cfrac%7B5%7D%7B2%7D%20%29%5E2%2B5%28-%5Cfrac%7B5%7D%7B2%7D%20%29%2B4%5C%5Cy_v%3D%5Cfrac%7B25%7D%7B4%7D%20-%5Cfrac%7B25%7D%7B2%7D%20%2B4%5C%5C%5C%5Cy_v%3D%5Cfrac%7B25%7D%7B4%7D%20-%5Cfrac%7B50%7D%7B4%7D%2B%5Cfrac%7B16%7D%7B4%7D%20%5C%5Cy_v%3D-%5Cfrac%7B9%7D%7B4%7D)
This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).
We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the
. Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.
When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):
![f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0](https://tex.z-dn.net/?f=f%28-1%29%20%3D%20%28-1%29%5E2%2B5%28-1%29%2B4%3D%201-5%2B4%20%3D%200%5C%5Cf%28-4%29%3D%28-4%29%5E2%2B5%28-4%29_4%3D16-20%2B4%3D0)
The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.
Then, the two roots are: x = -1 and x = -4.