Question

Can someone show me how I would solve this using the Quadratic Formula? 2x^2 – 16x + 32 = 0

Answer 1

do you know the quadratic form?

Answer 2

I think x=4

Answer 3

Compare your quadratic equation with \(ax^2+bx+c=0\)
find a,b,c
then the two roots of x are:
\(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

Answer 4

The +- part always confuses me. Could you demonstrate how I'd substitute an equation like this into the form?

Answer 5

\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \text{ means:} \\
\quad x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \text{, and} \\
\quad x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} \text{ are the two solutions} \]
Basically, its a more compact way of writing both solutions.
The typical method is to substitute your values, simplify your sqrt(b^2 - 4ac) and -b, and then deal with splitting the \(\pm\) into two solutions + and -.

Answer 6

2x^2-8x-8x+32=0
2x(x-4)-8(x-4)=0
(2x-8)(x-4)=0
x=4,2

Answer 7

Can I ask why you wrote it as 2x^2-8x-8x+32=0

Answer 8

its supposed to be b+-4ac?