3x^2-2x-2>0?

Question

I entered my last question incorrectly. How would you solve this inequality?

Answer 1

3x^2 - 2x - 2 > 0

There's a traditional method, but I'm going to show you an alternate one. Complete the square. This is using the fact that sqrt(y^2) = |y|, and that, for absolute inequalities:

|y| < a translates to (-a < y < a) for positive a value, and
|y| > a translates to y > a OR y < -a.

3(x^2 - (2/3)x) - 2 > 0

3(x^2 - (2/3)x + 1/9) - 2 - 3/9 > 0
3(x - (1/3))^2 - 2 - 1/3 > 0

3(x - (1/3))^2 - 6/3 - 1/3 > 0

3(x - (1/3))^2 - 7/3 > 0
3(x - (1/3))^2 > 7/3

(x - (1/3))^2 > 7/9

Take the square root of both sides and use that property I mentioned.

|x - (1/3)| > sqrt(7/9)

|x - (1/3)| > sqrt(7)/3

This translates as

x - (1/3) > sqrt(7)/3 OR x - (1/3) < -sqrt(7)/3

x > (1/3) + sqrt(7)/3 OR x < (1/3) - sqrt(7)/3

x > [1 + sqrt(7)]/3 OR x < [1 - sqrt(7)]/3

Answer 2

You need to find the critical values.

Set the 3x^2-2x-2 equal to zero and solve for x.
This will give you [1+/- sqrt(7)]/3

This seperates the number line into three regions. Test a value from each region to determine if
x < [1- sqrt(7)]/3, [1- sqrt(7)]/3< x < [1+ sqrt(7)]/3, or x >[1+ sqrt(7)]/3

Answer 3

3x²-2x-2>0
d = -2² - 4.3.-2
d = 4 + 24 = 28

x = (-2 +/- \/28) : 2
x' = (-2 + 2\/7) : 2
x' = -1 + \/7
x" = -1 - \/7
Solution: {x belongs to R | "x" different of -1 + \/7 and -1 - \/7}