4.4: Quadratic Formula Starter

4.4: Quadratic Formula Starter

Multiple Choice

If we are going to solve the following equation using the quadratic formula we first need to get everything to ones side of the equation.  What do you get when you move everything to the left side of the equation?

 $2x^2+4x+8=5-4x$  

•  $2x^2+13=0$  

•  $2x^2+8x+3=0$  

•  $2x^2+8x+13=0$  

•  $2x^2+3=0$  

Multiple Choice

Now that we have everything on one side, what will it look like when we substitute the a, b, and c values into the quadratic formula?

 $2x^2+8x+3=0$ 

•  $x=\frac{8\pm\sqrt{8^2-4\left(2\right)\left(3\right)}}{2\left(2\right)}$  

•  $x=\frac{-8\pm\sqrt{8^2+4\left(2\right)\left(3\right)}}{2\left(2\right)}$  

•  $x=\frac{-8\pm\sqrt{8^2-4\left(2\right)\left(3\right)}}{2\left(2\right)}$  

Multiple Choice

Which of the following equations should we get after we simplify the denominator and the expression under the square root?

 $x=\frac{-8\pm\sqrt{8^2-4\left(2\right)\left(3\right)}}{2\left(2\right)}$  

•  $x=\frac{-8\pm\sqrt{40}}{4}$  

•  $x=\frac{-8\pm\sqrt{360}}{4}$  

•  $x=\frac{-8\pm\sqrt{-8}}{4}$  

•  $x=\frac{-8\pm\sqrt{48}}{4}$  

Multiple Choice

What do you get when you simplify the following answer?


 $x=\frac{-8\pm\sqrt{40}}{4}$  

•  $x=-2\pm\sqrt{10}$  

•  $x=-2\pm2\sqrt{10}$  

•  $x=\frac{-4\pm\sqrt{10}}{2}$  

•  $x=-\frac{3\sqrt{10}}{2}\ or\ \ x=-\frac{5\sqrt{10}}{2}$