+0  
 
0
36
5
avatar+846 

Hi friends,

I have a few questions tonight, hope you do not mind...

 

calculate x

\((x+2)(2x-3)\leq-3\)

 

I did this:

\(x+2\leq-3\)   or   \(2x-3\leq-3\)

\(x\leq-5\)  or  \(x\leq0\)

 

Does not seem right??

 Jun 2, 2021
 #1
avatar+368 
+3

Hi there, 

i cant see ur steps but here are mine:

\(\left(x+2\right)\left(2x-3\right)\le-3\)

\(2x^2+x-3\le0\)

\(\left(x-1\right)\left(2x+3\right)\le0\)           (u get this by factoring \((x-1)(2x+3)\))

\(-\frac{3}{2}\le x\le 1\)                          (identify the interval)

 

More here: https://www.symbolab.com/solver/step-by-step/%5Cleft(x%2B2%5Cright)%5Cleft(2x-3%5Cright)%5Cle-3

 Jun 2, 2021
 #3
avatar+846 
+2

Hi mworkhard222,

 

thank you for the quick response...but please just tell me if you don't mind...

 

I thought the factors were already out there...meaning the (x+2) and the (2x-3)??

juriemagic  Jun 2, 2021
 #4
avatar+846 
+2

Hi mworkhard222,

I've got it...thank you very much..

juriemagic  Jun 2, 2021
 #2
avatar+1780 
+2

(x+2)(2x-3) =< -3

2x^2 + x - 6 =< -3

2x^2 + x - 3 =< 0

(2x+3)(x-1) =< 0 

 

If x > 1, both 2x-1 and x+1 will be positive causing x to not satisfy the equation. positive * positive = positive

If x < -1.5, both 2x-1 and x+1 will be negative causing x to not satisfy the equation. negative * negative = positive

So x has to be inbetween, x = [-1.5, 1]

 

=^._.^=

 Jun 2, 2021
 #5
avatar+846 
+2

Hi catmg,

I've got it thanx, my mistake was I thought the first equation already had the factors, but the factors are really only established once the equation is equal to, or more, or less than 0..and not -3.  Thank you kindly.

juriemagic  Jun 2, 2021
edited by juriemagic  Jun 2, 2021

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