Q. \(x^2 = 50\)

Q: How do you simplify \sqrt{50}?

\sqrt{50}=\sqrt{25\cdot 2}=5\sqrt{2}.

Q: Is there a solution with x=0?

No. If x=0, then x^2=0\neq 50. The solutions are only x=\pm 5\sqrt{2}.

Q: What if I take the square root incorrectly?

From x^2=50, you must use x=\pm\sqrt{50}, not just x=\sqrt{50}.

Q: How do you solve x^2=50 using factoring?

Rewrite as x^2-50=0. Over reals, factoring isn’t clean; use square roots: x=\pm\sqrt{50}=\pm 5\sqrt{2}.

Q: What are the approximate decimal solutions?

5\sqrt{2}\approx 5(1.4142)\approx 7.071. So x\approx \pm 7.071.

Q: What are the complex solutions?

Since 50>0, there are no nonreal solutions. Solutions stay real: x=\pm 5\sqrt{2}.

Answer

We solve the equation \(x^2=50\) by taking square roots of both sides:

\[
x=\pm \sqrt{50}=\pm \sqrt{25\cdot 2}=\pm 5\sqrt{2}
\]

Final result: \(x=5\sqrt{2}\) or \(x=-5\sqrt{2}\).

Detailed Explanation

We are asked to solve the equation

\[
x^2=50
\]

Step 1: Take the square root of both sides

Because \(x^2\) is a square, \(x\) can be positive or negative. So we take the square root of both sides:

\[
\sqrt{x^2}=\sqrt{50}
\]

Since \(\sqrt{x^2}=|x|\), we get:

\[
|x|=\sqrt{50}
\]

Step 2: Solve for \(x\) using the absolute value

The equation \(|x|=\sqrt{50}\) means \(x\) can be either

\[
x=\sqrt{50}
\quad \text{or} \quad
x=-\sqrt{50}
\]

Step 3: Simplify \(\sqrt{50}\)

Factor \(50\) as \(25\cdot 2\):

\[
\sqrt{50}=\sqrt{25\cdot 2}
\]

Use \(\sqrt{25}=5\):

\[
\sqrt{50}=5\sqrt{2}
\]

Final Answer

\[
x=5\sqrt{2}
\quad \text{or} \quad
x=-5\sqrt{2}
\]

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Algebra FAQ

What are all solutions to \(x^2=50\)?

\(x=\pm\sqrt{50}=\pm 5\sqrt{2}\).

How do you simplify \(\sqrt{50}\)?

\(\sqrt{50}=\sqrt{25\cdot 2}=5\sqrt{2}\).

Is there a solution with \(x=0\)?

No. If \(x=0\), then \(x^2=0\neq 50\). The solutions are only \(x=\pm 5\sqrt{2}\).

What if I take the square root incorrectly?

From \(x^2=50\), you must use \(x=\pm\sqrt{50}\), not just \(x=\sqrt{50}\).

How do you solve \(x^2=50\) using factoring?

Rewrite as \(x^2-50=0\). Over reals, factoring isn’t clean; use square roots: \(x=\pm\sqrt{50}=\pm 5\sqrt{2}\).

What are the approximate decimal solutions?

\(5\sqrt{2}\approx 5(1.4142)\approx 7.071\). So \(x\approx \pm 7.071\).

What are the complex solutions?

Since \(50>0\), there are no nonreal solutions. Solutions stay real: \(x=\pm 5\sqrt{2}\).

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