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Find the zeros of the function: f(x) x^2-4x√5 +19
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guess you can use the quadratic formula
it is really \[f(x)=x^2-4\sqrt{5}x+19\]?
It says \[f(x)=x^2-4x \sqrt{5}+19\]
same thing
use \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]with \[a=1, b=-4\sqrt5, c=19\]
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I got \[\frac{ 4\sqrt{5}\pm2 }{ 2 }\]
Do I have to make it \[2\sqrt{5}\pm1\] instead?
Yes, good job!
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