# When I solve the inquality 2x^2 - 6 < 0,

I get x < + or - sqrt(3)

So how do I write the solution?
Is it (+sqrt(3),-sqrt(3)) or (-infinity, sqrt(3))?
Why?

Thanks.

So would this work?

abs x < ( sqrt 3 )

or
- sqrt 3<x<sqrt 3

I don't know the reason

fuji

## To solve the inequality 2x^2 - 6 < 0, follow these steps:

1. Start by solving the corresponding equation, 2x^2 - 6 = 0, by setting it equal to zero and factoring or using the quadratic formula. In this case, we can factor out a 2 and rewrite the equation as 2(x^2 - 3) = 0.

2. Now, set each factor equal to zero and solve for x:
x^2 - 3 = 0
x^2 = 3
x = ±√3

3. This gives us two solutions, x = √3 and x = -√3.

To write the solution set, we need to determine which values of x satisfy the original inequality 2x^2 - 6 < 0.

4. We can test different values of x in each interval created by the critical points (√3 and -√3) to determine where the inequality holds true.

a) Choose a value less than -√3, such as -4, and substitute it into the inequality:
2(-4)^2 - 6 < 0
32 - 6 < 0
26 < 0

Since 26 is not less than zero, this interval does not satisfy the inequality.

b) Choose a value greater than -√3 and less than √3, such as 0, and substitute it into the inequality:
2(0)^2 - 6 < 0
-6 < 0

Since -6 is less than zero, this interval satisfies the inequality.

c) Choose a value greater than √3, such as 4, and substitute it into the inequality:
2(4)^2 - 6 < 0
32 - 6 < 0
26 < 0

Similar to interval a), 26 is not less than zero, so this interval does not satisfy the inequality.

Based on the above analysis, we can conclude that the solution to the inequality 2x^2 - 6 < 0 is (-√3, √3).