## mathmath333 one year ago Find the range of $$x$$

1. mathmath333

\large \color{black}{\begin{align} (2x+3)(x-5)<0\hspace{.33em}\\~\\ \end{align}}

2. ganeshie8

the parabola cuts the x axis at points -3/2 and 5 so it looks like this |dw:1433764773998:dw|

3. ganeshie8

$-\frac{3}{2}~\lt~x\lt~5$

4. mathmath333

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 $$x<5$$ ?

5. ganeshie8

product of two negatives always gives you > 0

6. ganeshie8

$a*b \lt 0$ implies $a\lt 0,~\text{and}~ b\gt 0$ or $a\gt 0,~\text{and}~b\lt 0$

7. ganeshie8

$(2x+3)*(x-5) \lt 0$ implies $(2x+3)\lt 0,~\text{and}~ (x-5)\gt 0$ or $(2x+3)\gt 0,~\text{and}~(x-5)\lt 0$

8. mathmath333

so by the above as u stated it can also be $$x<-\dfrac{3}{2}$$ and $$x>5$$

9. ganeshie8

Yes, but are there any numbers in real that are both less than -3/2 and greater than 5 ?

10. mathmath333

no

11. ganeshie8

so that case yields an emptyset

12. ganeshie8

we work the other case

13. mathmath333

is that the only algebraic method u stated

14. ganeshie8

i don't like algebraic method when the product has only 2 factors i prefer visualizing a parabola

15. ganeshie8

because the x intercepts are readily avilable when you're given the inequality in product form

16. mathmath333

|dw:1433765981235:dw| how did u visualize the shaded region

17. ganeshie8

\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

18. ganeshie8
19. anonymous

@mathmath333 that isnt you in your profile picture :)

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