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i need to find 2 distinct numbers where (x+3)/(x^2+9)=1/6
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We have\[\frac{x+3}{x^2+9}=\frac{1}{6}\] cross multiplying we get \[ 6x+18=x^2+9\] Bring all the terms to the same side \[x^2-6x-9=0\] Can you solve it now using quadratic formula?
i believe so. 6+/-square root 36-4(1)(-9)/2
yeah, correct :D
huzzah
confused though
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Why?
what would the answer be? 6/-(36-36)/2?
6+/-(0)/2
\[x =\frac{6\pm\sqrt{36-4 \times (1)\times (-9) }}{2}\] we get \[x =\frac{6\pm\sqrt{72 }}{2}\] 72=36*2 so \[x =\frac{6\pm6\sqrt{2 }}{2}\] \[x=3\pm 3\sqrt{2}\] or \[x=3(1\pm \sqrt{2})\]
my mistake, thanks ash
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