How do you find the zeros, real and imaginary, of $y=2(x-3)^2$ using the quadratic formula?

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Answer 1 · 2016-07-13T09:50:21

$x=3$

Squaring the bracket gives:

$y=2(x^2-6x+9)$

$y=2x^2-12x+18$
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Using $y=ax^2+bx+c$ where

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

$a=2"; "b=-12"; "c=18$
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$x=(+12+-sqrt((-12)^2-4(2)(18)))/(2(2))$

$x=(+12+-sqrt(144-144))/4$

$x=+3" "$ thus the x-axis is tangential to the vertex

Tony B

How do you find the zeros, real and imaginary, of y=2(x-3)^2 y=2(x-3)^2 using the quadratic formula?