Question

Solve the following system algebraically.y = x2 – 20y = x – 8A) (3,–4) and (–11,2)B) (5,2) and (5,8)C) (0,7) and (10,–7)D) (–3,–11) and (4,–4)

Answer 1

We have the system

$\begin{gathered} y=x^2-20 \\ y=x-8 \end{gathered}$

So we can equalize the two equations to get x

$\begin{gathered} x^2-20=x-8 \\ x^2-20+8-x=0 \\ x^2-x-12=0 \end{gathered}$

So now we can solve the quadratic equation to find x, we are going to use the quadratic formula

$\frac{-b±√\left(b²-4ac\right)}{2a}$

We have a = 1, b= - 1 and c = -12, so replacing we get

$\frac{-(-1)±√\left((-1)²-4\cdot1\cdot(-12)\right)}{2\cdot(1)}$

this is

$\frac{1\pm\sqrt{1+48}}{2}=\frac{1\pm\sqrt{49}}{2}=\frac{1\pm7}{2}$

Then the solutions for x are x = (1 - 7)/2 = -6/2 = -3, and x = (1+7)/2 = 8/2 = 4. Now we replace in the second equation and we get two solutions for y, using x = -3 we get y = -3 - 8 = -11, and using x = 4 we get y = 4 - 8 = -4. So the two points are: (-3,-11) and (4,-4). THE ANSWER IS D.