Q. \(x^2 = 20\)

Q: Are there complex solutions to x^2=20?

Yes, but 20>0, so the solutions are already real: \[x=\pm 2\sqrt{5}.\] No additional non-real roots occur.

Q: How do you solve x^2=20 using square roots carefully?

Take square roots: \[x=\pm\sqrt{20}.\] The \pm appears because squaring removes sign information.

Q: Check the solutions by substitution.

For x=2\sqrt{5}: (2\sqrt{5})^2=4\cdot 5=20. For x=-2\sqrt{5}: (-2\sqrt{5})^2=20.

Q: Solve x^2=20 but approximate decimal values.

\sqrt{20}\approx 4.4721, so \[x\approx \pm 4.4721.\]

Answer

We solve \(x^2 = 20\). Taking square roots gives \(x = \pm \sqrt{20} = \pm \sqrt{4\cdot 5} = \pm 2\sqrt{5}\).

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

Detailed Explanation

We are given the equation \(x^2 = 20\). The goal is to solve for \(x\).

Step 1: Take the square root of both sides.

Since \(x^2 = 20\), taking the square root of both sides gives:
\[
x = \pm \sqrt{20}.
\]

Step 2: Simplify \(\sqrt{20}\).

Factor \(20\) as \(4 \cdot 5\):
\[
\sqrt{20} = \sqrt{4 \cdot 5}.
\]

Use the square root property \(\sqrt{a \cdot b} = \sqrt{a}\sqrt{b}\):
\[
\sqrt{4 \cdot 5} = \sqrt{4}\sqrt{5}.
\]

Since \(\sqrt{4} = 2\), this becomes:
\[
\sqrt{20} = 2\sqrt{5}.
\]

Step 3: Substitute back into the solution.

So,
\[
x = \pm 2\sqrt{5}.
\]

Final Answer:

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

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Algebra FAQ

Solve \(x^2=20\).

\[x=\pm\sqrt{20}=\pm 2\sqrt{5}.\]

What are the real solutions to \(x^2=20\)?

\[x=\pm 2\sqrt{5}.\] Both are real numbers.

Are there complex solutions to \(x^2=20\)?

Yes, but \(20>0\), so the solutions are already real: \[x=\pm 2\sqrt{5}.\] No additional non-real roots occur.

How do you simplify \(\sqrt{20}\) in \(x=\pm\sqrt{20}\)?

\[\sqrt{20}=\sqrt{4\cdot 5}=2\sqrt{5}.\]

How do you solve \(x^2=20\) using square roots carefully?

Take square roots: \[x=\pm\sqrt{20}.\] The \(\pm\) appears because squaring removes sign information.

Check the solutions by substitution.

For \(x=2\sqrt{5}\): \((2\sqrt{5})^2=4\cdot 5=20.\) For \(x=-2\sqrt{5}\): \((-2\sqrt{5})^2=20.\)

Solve \(x^2=20\) but approximate decimal values.

\(\sqrt{20}\approx 4.4721\), so \[x\approx \pm 4.4721.\]

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