Hello, Lightning!
The quadratic formula is a formula that solves for the roots of any quadratic. Let's apply it to the quadratic \(4x^2+7x+k\).
\(a=4, b=7, c=k;\\ x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) | Substitute in the appropriate values into the formula. |
\(x_{1,2}=\frac{-7\pm\sqrt{7^2-4*4*k}}{2*4} \) | Simplify. |
\(x_{1,2}=\frac{-7\pm\sqrt{49-16k}}{8}\) | |
Obviously, we do not know what k is, but we do know that the roots of the quadratic with the unknown k are \(x_{1,2}=\frac{-7\pm i\sqrt{15}}{8}\).
Do you think you can take it on from here?
The reason Gavin did it his way is that most times this would not work for a table of values.
You would not be able to pick the exact right x value to put into the table. BUT this one has been chosen so that it is easy.
f(x)=x^2-6x+8 this is a parabola, maybe you are meant to know that
g(x)=x-2 this is a line and maybe you are meant to know that as well.
There will be at most two points of intersection between a line and a parabola, lets see if we can find them
f(x)=x^2-6x+8
x | 0 | 1 | 2 | 3 | 4 | 5 |
f(x) | 0-0+8=8 | 1-6+8=3 | 2^2-6*2+8 =4-12+8 =0 | etc |
g(x)=x-2
x | 0 | 1 | 2 | 3 | 4 | 5 |
g(x) | 0-2=2 | 1-2=-1 | 2-2=0 | etc |
I can see that for x=2 f(2)=0 and g(2)=0
so f(x)=g(x) when x=2 So the first solution is x=2
Now continue the tables and see if you can find another match :)