Question 5
To use the quadratic formula, we need to rewrite the equation so that all the terms are on one side: by subtracting (x+13) from both sides, we get that [tex]x^2-x-6=0[/tex]. The quadratic formula tells us that the roots of this are [tex]\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex], where a=1, b=-1, and c=-6. This is equal to [tex]\frac{-(-1)\pm \sqrt{(-1)^2-4(1)(-6)}}{2(1)} = \frac{1 \pm 5}{2}[/tex], which gives us the two roots -2 and 3.
Question 7
If you look at the attached graph, you will see that the graph of x^2-x-6 intersects the x-axis at the points (-2,0) and (3,0).
Question 8
The roots are the same. Both finding the x-intercepts and solving for the roots of a quadratic using the quadratic formula will give the values of x for which the function value is zero.
