36But the answer is different....!!
how to take the intersection of these three inequalities
1> 6x^2 - 5x + 1 < 0
2> 3x^2 + 2x - 1 > 0
3> 2x^2 + x - 3 < 0
please help i m muddledersection of these three inequalities
1> 6x^2 - 5x + 1 < 0
2> 3x^2 + 2x - 1 > 0
3> 2x^2 + x - 3 < 0
please help i m muddled
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2 Answers
306x^2-5x+1<0 \Rightarrow \left(2x-1 \right)\left(3x-1 \right)<0
\Rightarrow x\in \left(\frac{1}{3}, \frac{1}{2}\right)
3x^2+2x-1>0 \Rightarrow \left(x+1 \right)\left(3x-1 \right)
\Rightarrow x\in \left(-\infty,-1 \right)\cup \left(\frac{1}{3},\infty \right)
2x^2+x-3<0 \Rightarrow \left(2x+3 \right)\left(x-1 \right)<0
\Rightarrow x\in \left(-\frac{3}{2},1 \right)
Now, the required intersection is:
\left(\frac{1}{3},\frac{1}{2}\right)\cap\left(\left(-\infty,-1 \right)\cup \left(\frac{1}{3},\infty \right) \right)\cap\left(-\frac{3}{2},1 \right)=\left(\frac{1}{3},\frac{1}{2} \right)