Find the range of \(x\)

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Find the range of \(x\)

Mathematics
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\(\large \color{black}{\begin{align} (2x+3)(x-5)<0\hspace{.33em}\\~\\ \end{align}}\)
the parabola cuts the x axis at points -3/2 and 5 so it looks like this |dw:1433764773998:dw|
\[-\frac{3}{2}~\lt~x\lt~5\]

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But if i evaluate individually \((2x+3)<0\) and \((x-5)<0\) then it comes as \(x<-\dfrac{3}{2}\) and \(x<5\) so it should be \(x<5\) ?
product of two negatives always gives you > 0
\[a*b \lt 0 \] implies \[a\lt 0,~\text{and}~ b\gt 0\] or \[a\gt 0,~\text{and}~b\lt 0\]
\[(2x+3)*(x-5) \lt 0 \] implies \[(2x+3)\lt 0,~\text{and}~ (x-5)\gt 0\] or \[(2x+3)\gt 0,~\text{and}~(x-5)\lt 0\]
so by the above as u stated it can also be \(x<-\dfrac{3}{2}\) and \(x>5\)
Yes, but are there any numbers in real that are both less than -3/2 and greater than 5 ?
no
so that case yields an emptyset
we work the other case
is that the only algebraic method u stated
i don't like algebraic method when the product has only 2 factors i prefer visualizing a parabola
because the x intercepts are readily avilable when you're given the inequality in product form
|dw:1433765981235:dw| how did u visualize the shaded region
\[\large \color{black}{\begin{align} (2x+3)(x-5)\color{red}{<}0\hspace{.33em}\\~\\ \end{align}}\] you want the product to be less than 0 so just shade the region below x axis
check this https://www.khanacademy.org/math/algebra2/polynomial_and_rational/quad_ineq/v/quadratic-inequalities-visual-explanation
@mathmath333 that isnt you in your profile picture :)

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