How do you find the zeros of $y = -2x^2 + 8x -2$ using the quadratic formula?

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Answer 1 · 2018-03-15T22:32:00

$x=2+-sqrt3$

The Quadratic Formula states that if a quadratic is of the form $ax^2+bx+c$ , it's zeroes are at

$(-b+-sqrt(b^2-4ac))/(2a)$

Our quadratic is in standard form, therefore we know

$a=-2$
$b=8$
$c=-2$

Now, we can just plug in. We get:

$(-8+-sqrt(8^2-4(-2)(-2)))/(2(-2)$

Which simplifies to

$(-8+-sqrt(64-16))/-4$

Which can be further simplified to

$(-8+-sqrt48)/-4$

At this point, we can factor $sqrt48$ into $color(blue)(sqrt16*sqrt3)$ , because $sqrt(ab)=sqrta*sqrtb$ . This simplifies to

$(-8+-color(blue)(4sqrt3))/-4$

We can factor out a $4$ because all of the terms have a $4$ in common. We get:

$(-2+-sqrt3)/-1$

Which can be simplified as

$2+-sqrt3$

How do you find the zeros of y = -2x^2 + 8x -2 y = -2x^2 + 8x -2 using the quadratic formula?