Question
Solve using formula.
$y^2 + \frac{1}{3}y = 2$
$y^2 + \frac{1}{3}y = 2$
✓
Answer
$3 y^2+y-6=0 $
$\Rightarrow 3 y^2+y-6=0 \text { compare with } a x^2+b x+c=0 $
$ \Rightarrow a=3, b=1 \text { and } c=-6 $
$ \therefore b^2-4 a c=1^2-4(3)(-6) $
$ =1+72$
$ =73$
$ Y=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$ \Rightarrow y=\frac{-1 \pm \sqrt{73}}{2 \times 3} $
$ \Rightarrow y=\frac{-1 \pm \sqrt{73}}{6} $
$\Rightarrow y=\frac{-1+\sqrt{73}}{6} \text { or } y=\frac{-1-\sqrt{73}}{6}$
$\Rightarrow 3 y^2+y-6=0 \text { compare with } a x^2+b x+c=0 $
$ \Rightarrow a=3, b=1 \text { and } c=-6 $
$ \therefore b^2-4 a c=1^2-4(3)(-6) $
$ =1+72$
$ =73$
$ Y=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$ \Rightarrow y=\frac{-1 \pm \sqrt{73}}{2 \times 3} $
$ \Rightarrow y=\frac{-1 \pm \sqrt{73}}{6} $
$\Rightarrow y=\frac{-1+\sqrt{73}}{6} \text { or } y=\frac{-1-\sqrt{73}}{6}$
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$3-5$
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$7-9$
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Frequency
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