trying to solve the inequality x^2-2x-2>0?

Answer 1

x^2 - 2x - 2 > 0

One method is to complete the square, and then use the fact that sqrt(y^2) = |y|.

x^2 - 2x + 1 - 2 - 1 > 0

(x - 1)^2 - 3 > 0

(x - 1)^2 > 3

Taking the square root of both sides,

|x - 1| > sqrt(3)

Absolute inequalities in the from |y| > c translate into
y > c OR y < -c. Therefore

x - 1 > sqrt(3), so x > 1 + sqrt(3).

x - 1 < -sqrt(3), so x < 1 - sqrt(3)

Therefore, the solution set is

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

Answer 2

sparkling up the inequality: x2 - 5 > 0? first of all, the question could particularly say 2x no longer x2. 2x - 5 > 0 Secondly, carry 5 to the different area of the inequality. So, -5 will become +5. Do you spot why? we've: 2x > 5 next: Divide the two aspects of the inequality by skill of two to sparkling up for x. very final answer: x > 5/2. What does our answer advise? It skill that despite the fee of x is, it is greater than the fraction 5/2. i'm hoping this facilitates. Guido P.S. shop this concepts: whilst multiplying or dividing by skill of a detrimental variety, you could replace the process the examine on your very final answer.

Answer 3

Solve for boundary points first with the quadratic formula,
x = 1±√3

Since y = x^2-2x-2 opens up, the solution of the inequality is,
x < 1-√3
or
x > 1+√3

Answer 4

You need to find the critical values.

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

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

Answer 5

The easiest way is to sketch the quadratic f(x) = x^2 - 2x - 2, and see where f(x) > 0. Use calculus to find the turning point and the roots of the equation will tell you the two x - intercepts. These two roots will determine your answer.

If you don't know calculus then I can tell you another way ...

Answer 6

discriminant is=4+4*2 =12
x1=(2-2sqrt3)/2
x2=(2+2sqrt3)/2 so the inequality can be write
(x-x1)(x-x2)>0
x -∞ x1 x2 +∞
x-x1 minus 0 plus
x-x2 minus 0 plus
(x-x1)(x-x2) plus 0 minus 0 plus

so x^2-2x-2>0 for x ( -∞, (2-2sqrt3)/2) U((2+2sqrt3)/2,+∞)

Answer 7

x²-2x-2>0
d = -2² - 4.1.-2
d = 4 + 8 = 12

x = (-2 +/- \/12) : 2
x' = (-2 + 2\/3) : 2
x' = -1 + \/3
x" = -1 - \/3
Solution: {x belongs to R | "x" different of -1 + \/3 and -1 - \/3}