How do you solve $(3 x - 5) (x - 2) < 0$ $(3x - 5) ( x - 2 ) < 0$ ?

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Answer 1 · 2015-11-07T21:25:07

$x \in (\frac{5}{3}, 2)$ $x in (5/3,2)$

$(3 x - 5) (x - 2)$ $(3x-5)(x-2)$ is the equation of a parabola
with x-intercepts at $\frac{5}{3}$ $5/3$ and $2$ $2$

If expanded into standard form the leading term is $3 x^{2}$ $3x^2$

$\Rightarrow$ $rArr$ the leading term coefficient is positive

$\Rightarrow$ $rArr$ the parabola opens upward

$\Rightarrow$ $rArr$ the points between the x-intercepts are below the X-axis

$\Rightarrow$ $rArr$ the value of the parabola is $< 0$ $< 0$ when $x \in (\frac{5}{3}, 2)$ $x in (5/3, 2)$

How do you solve (3x - 5) ( x - 2 ) < 0 (3x−5)(x−2)<0(3x - 5) ( x - 2 ) < 0 ?