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   If $\mid 9y-6 \mid = 3$, then $y^{2}-\dfrac{4}{3}y$ is .

  1. $0$
  2. $+\dfrac{1}{3}$
  3. $-\dfrac{1}{3}$
  4. undefined
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2 Answers

Best answer
12 votes
12 votes
$\mid {9y-6} \mid =3$

squaring both side

$81y^{2}-108y+36=9$

dividing both side by $9$

$9y^{2}-12y+4=1$

again dividing by $9$

$y^{2}-\dfrac{4}{3}y+\dfrac{4}{9}=\dfrac{1}{9}$

$\Rightarrow y^{2}-\dfrac{4}{3}y=\dfrac{1}{9}-\dfrac{4}{9}=-\dfrac{1}{3}.$

Hence answer is option $C$
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5 votes
5 votes
|9y-6|=3

So 9y-6=-3 =>y=1/3

and 9y-6=3 =>y=1

putting y={1,1/3} in Y$^{2}$-4y/3 We get in both cases  -$\frac{1}{3}$

So C is the answer
Answer:

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