Q. \(x^2 = 100\)

Q: Solve x^2=100.

Take square roots: x=\pm\sqrt{100}=\pm 10.

Q: Why are there two solutions?

Squaring loses sign information, since ( -10)^2=(10)^2=100. So both x=10 and x=-10 satisfy the equation.

Q: What if the equation is x^2=a with a>0?

Then x=\pm\sqrt{a}. For a<0, there are no real solutions (but complex ones exist).

Q: How do I check the solutions?

Substitute: 10^2=100 and (-10)^2=100. Both work, so the solutions are correct.

Q: What is the general solution set for x^2=100?

The solution set is \{ -10,\,10 \}.

Q: Can I write x=\sqrt{100} without the negative?

Not fully. \sqrt{100}=10 is only the principal (nonnegative) root. The full solution requires x=\pm 10.

Answer

Given \(x^2 = 100\). Take the square root of both sides:

\[
x = \pm \sqrt{100} = \pm 10
\]

So the solutions are \(x = 10\) and \(x = -10\).

Detailed Explanation

We are asked to solve the equation

\[ x^2 = 100. \]

Step 1: Take the square root of both sides.

To undo squaring, apply the square root to each side. Remember that if \(x^2 = 100\), then \(x\) can be both the positive and negative square roots of \(100\).

\[
\sqrt{x^2} = \sqrt{100}.
\]

This gives

\[
x = \pm \sqrt{100}.
\]

Step 2: Compute \( \sqrt{100} \).

Since \(100 = 10^2\), we have

\[
\sqrt{100} = 10.
\]

Step 3: Write the two solutions.

Substitute \(10\) back into \(x = \pm \sqrt{100}\):

\[
x = \pm 10.
\]

Final Answer:

The solutions to \(x^2 = 100\) are

\[
x = 10 \quad \text{and} \quad x = -10.
\]

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Algebra FAQ

Solve \(x^2=100\).

Take square roots: \(x=\pm\sqrt{100}=\pm 10\).

Why are there two solutions?

Squaring loses sign information, since \(( -10)^2=(10)^2=100\). So both \(x=10\) and \(x=-10\) satisfy the equation.

What if the equation is \(x^2=a\) with \(a>0\)?

Then \(x=\pm\sqrt{a}\). For \(a<0\), there are no real solutions (but complex ones exist).

How do I check the solutions?

Substitute: \(10^2=100\) and \((-10)^2=100\). Both work, so the solutions are correct.

What is the general solution set for \(x^2=100\)?

The solution set is \(\{ -10,\,10 \}\).

Can I write \(x=\sqrt{100}\) without the negative?

Not fully. \(\sqrt{100}=10\) is only the principal (nonnegative) root. The full solution requires \(x=\pm 10\).

Solve x²=100 step by step.
Check your answer and units.

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