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how do you find the zeros in

y=-6x^2 + 42x -6

Guest Mar 29, 2017
 #1
avatar+92625 
+2

Use the quadratic formula

 

https://www.youtube.com/watch?v=O8ezDEk3qCg

 

a=-6, b=42, c=-6

Melody  Mar 29, 2017
 #2
avatar+13 
+1

\(y= -6x^2+42x -6\)

 

First, factor out a -6

\(y = -6(x^2 - 7x + 1)\)

 

Then, set the value in the parentheses equal to 0

\(x^2-7x+1=0\)

 

Now, use the quadratic formula to solve

\(y = {(7) \pm \sqrt{(7)^2-4(1)(1)} \over 2(1)}\)

\(y = {7 \pm 3 \sqrt5 \over 2}\)

\(y \approx 6.854, 0.146\)

 

Your zeros are approximately 6.854, and 0.146

austinruffino  Mar 29, 2017

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