+0

# Calculate X

0
105
5
+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
+502
+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)

Jun 2, 2021
#3
+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
+846
+2

Hi mworkhard222,

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

juriemagic  Jun 2, 2021
#2
+2205
+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
+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