Question

Need help on how to solve inequalities. I know how to do them graphing, but how do I do it mathematically?

1. \[x^2 + 2x - 3 \le 0\]
2. \[\frac{2x-1}{3x-2}\le1\]
3. \[x^2 + x + 1 > 0\]

Answer 1

@ajprincess @PaxPolaris @KingGeorge @dpaInc can you help me?

Answer 2

I do it by completing the square first. That is:
x^2 + 2x -3 <= 0 <=> (x^2 + 2x + 1) - 3 - 1 <= 0 <=> (x+1)^2 <= 4
Now take the square root to get: |x+1| <= 2
There are 2 cases:
i) If x+1 >= 0, then the modulus is: |x+1| = x+1 and we have x+1<=2 or x<=1.
ii) if x+1 < 0, then |x+1| = -(x+1) and we have -(x+1) <=2 or x>= -3.
So the solution will be: -3 <= x <= 1.

Answer 3

No. 3 is easier to solve:
x^2 + 2(1/2)x + (1/2)^2 - (1/2)^2 + 1 > 0 <=> (x+1/2)^2 > -3/4
The square on the left side is always positive so this inequality holds for all x.

Answer 4

1.
\[\Large x^2+2x-3 \le 0\]\[\implies \Large \left( x+3 \right)\left( x-1 \right) \le 0\]

you can see that the graph of y=(x+3)(x-1) will look like this:
so, from the graph \[\Large -3 \le x \le 1\]

Answer 5

even without graphing this:

for the product (x+3)(x-1) to be negative
one of them has to be negative and the other positive

so, \(\large x +3 \ge 0\), and \(\large x -1 \le 0\)

Answer 6

and 2.
\[\large {2x-1 \over 3x-2} \le 1\]
\[\implies \Large 2x-1 \le 3x-2\ \dots\ \left\{ \text{ where } 3x-2 \ne0\right\}\]