+1124 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??
+514 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
+1124 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)??
+2401 (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]
=^._.^=
+1124 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.
