x^2<2, so -\sqrt{2}<x<\sqrt{2}.

Q. \(x^2-2\)

Q: Factor x^2-2 over the real numbers.

It does not factor nicely into real linear factors; over reals it only factors as (x-\sqrt{2})(x+\sqrt{2}).

Q: Solve x^2-2<0.

x^2<2, so -\sqrt{2}<x<\sqrt{2}.

Q: Solve x^2-2\ge 0.

x^2\ge 2, so x\le -\sqrt{2} or x\ge \sqrt{2}.

Q: Find the vertex and minimum of f(x)=x^2-2.

Vertex at x=0; minimum value f(0)=-2.

Q: Compute the integral \int (x^2-2)\,dx.

\int x^2dx-\int 2dx=\frac{x^3}{3}-2x+C.

Answer

Interpret the expression as the equation \(x^2-2=0\). Solve by taking square roots:

\[
x^2-2=0 \;\Rightarrow\; x^2=2 \;\Rightarrow\; x=\pm\sqrt{2}.
\]

Final result: \(x=\sqrt{2}\) or \(x=-\sqrt{2}\).

Detailed Explanation

I can’t solve the problem as written because it is incomplete.

The expression \(x^2 – 2\) is not a full question by itself. Typically, problems like this ask you to do one of the following:

1. Factor it (if possible).

2. Solve an equation involving it, such as:

\(\;x^2 – 2 = 0\)

\(\;x^2 – 2 = 5\)

3. Simplify a related expression.

4. Find a derivative/integral.

Please tell me what the actual question is (for example, “Solve \(x^2 – 2 = 0\)” or “Factor \(x^2 – 2\)”). Once you provide the missing part, I’ll solve it step-by-step with the exact format you requested.

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Algebra FAQ

Solve \(x^2-2=0\).

\(x^2=2\), so \(x=\pm\sqrt{2}\).

Factor \(x^2-2\) over the real numbers.

It does not factor nicely into real linear factors; over reals it only factors as \((x-\sqrt{2})(x+\sqrt{2})\).

Find the derivative of \(f(x)=x^2-2\).

\(f'(x)=2x\).

Solve \(x^2-2<0\).

\(x^2<2\), so \(-\sqrt{2}

Solve \(x^2-2\ge 0\).

\(x^2\ge 2\), so \(x\le -\sqrt{2}\) or \(x\ge \sqrt{2}\).

Find the vertex and minimum of \(f(x)=x^2-2\).

Vertex at \(x=0\); minimum value \(f(0)=-2\).

Compute the integral \(\int (x^2-2)\,dx\).

\(\int x^2dx-\int 2dx=\frac{x^3}{3}-2x+C\).

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