Find all real numbers x such that x^2 -8>0. Write your final answer in interval notation.
I really just need a starter step. I remember these questions, but I'm not sure how to start. Please help!
x^2 - 8 > 0
x^2 > 8
That should help.
So the sign is flipped because we moved a negative number, right?
no, you can add and subtract anything without flipping the direction. Only if you multiply or divide by a negative value does the direction change.
It's really just a shortcut to moving things to the other side:
-2x < 8
0 < 8+2x
-8 < 2x
-4 < x
That's the same as dividing and flipping at once:
-2x < 8
x > -4
To solve the inequality \(x^2 - 8 > 0\), we need to find the values of \(x\) that make the expression \(x^2 - 8\) positive.
To start, we can solve for \(x\) as if the inequality were an equation:
\[x^2 - 8 = 0\]
To solve this equation, we can add 8 to both sides:
\[x^2 = 8\]
Next, we can take the square root of both sides of the equation. Remember that when we take the square root, we need to consider both the positive and negative square root:
\[x = \pm \sqrt{8}\]
Simplifying the square root of 8, we have:
\[x = \pm 2\sqrt{2}\]
Now, we can use the solutions obtained from the equation \(x^2 - 8 = 0\) and analyze the inequality \(x^2 - 8 > 0\).
Since \(x^2 - 8\) is a quadratic expression, it is equal to zero at \(x = \pm 2\sqrt{2}\), which means it changes sign at these points. Recall that when a quadratic expression changes sign, it goes from positive to negative or from negative to positive.
In this case, we want to find the intervals where \(x^2 - 8\) is greater than zero, which means we are looking for the values of \(x\) when the expression is positive. We can represent this as follows:
\((-\infty, -2\sqrt{2}) \cup (2\sqrt{2}, \infty)\)
This interval notation indicates that \(x\) can take any value less than \( -2\sqrt{2}\) or any value greater than \(2\sqrt{2}\) to satisfy the inequality \(x^2 - 8 > 0\).