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How do you find the zeros of: f(x) = 9x^2 - 27x + 20
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lets try this : substitute x= y + 3/2 see to it if you get it ..
Where did you get x=y+3/2 from?
Use the quadratic formula
-(-27) +_ sqrt(-27^2 -4(9)(20)/2(9)?
is that correct?
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yes
\[9x^{2}-27x+20=0\] \[x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\] \[x=\frac{27 \pm \sqrt{729-720}}{18}\] \[x=\frac{27 \pm \sqrt{9}}{18}\] \[x=\frac{27 \pm 3}{18}\]
...5 and 4 -5 and -4 -5/3 and -4/3 5/3and 4/3?
\[x=\frac{30}{18}\]or \[x=\frac{24}{18}\]
\[x=\frac{5}{3}\]or \[x=\frac{4}{3}\]
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