Q. \(x^2 = 81\)

Q: What values of \,x\, satisfy \,x^2=81\,?

x=\pm \sqrt{81}=\pm 9, so x=9 or x=-9.

Q: How do you solve \,x^2=81\, using square roots?

Take square root of both sides: \sqrt{x^2}=\sqrt{81}. Use principle: x=\pm 9.

Q: Why do we need both positive and negative solutions?

Squaring removes signs: both 9^2 and (-9)^2 equal 81. So both x=9 and x=-9 work.

Q: What is \,\sqrt{x^2}\, equal to?

\sqrt{x^2}=\lvert x\rvert. So \lvert x\rvert=9 gives x=9 or x=-9.

Q: How would you solve \,x^2-81=0\, by factoring?

Factor: (x-9)(x+9)=0. Then x-9=0 or x+9=0, so x=9 or x=-9.

Q: What are the solutions if \,x^2=0\, instead?

\sqrt{0}=0, so x^2=0 implies x=0 only (no \pm difference).

Q: Does \,x=\sqrt{81}\, always give all solutions?

No. \sqrt{81}=9 gives only x=9. The full set requires x=\pm 9.

Answer

We solve \(x^2=81\) by taking square roots on both sides.

\[
x=\pm\sqrt{81}=\pm 9
\]

Final result: \(x=9\) or \(x=-9\).

Detailed Explanation

We want to solve the equation \(x^2 = 81\).

Step 1: Take the square root of both sides.

Since \(x^2 = 81\), taking square roots gives:

\[
\sqrt{x^2} = \sqrt{81}
\]

So this becomes:

\[
|x| = 9
\]

Step 2: Convert the absolute value equation into two cases.

The equation \( |x| = 9 \) means that \(x\) can be positive or negative. So we have:

\[
x = 9 \quad \text{or} \quad x = -9
\]

Final Answer:

The solutions to \(x^2 = 81\) are:

\[
x = 9 \quad \text{and} \quad x = -9
\]

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Algebra FAQ

What values of \(\,x\,\) satisfy \(\,x^2=81\,\)?

\(x=\pm \sqrt{81}=\pm 9\), so \(x=9\) or \(x=-9\).

How do you solve \(\,x^2=81\,\) using square roots?

Take square root of both sides: \(\sqrt{x^2}=\sqrt{81}\). Use principle: \(x=\pm 9\).

Why do we need both positive and negative solutions?

Squaring removes signs: both \(9^2\) and \((-9)^2\) equal \(81\). So both \(x=9\) and \(x=-9\) work.

What is \(\,\sqrt{x^2}\,\) equal to?

\(\sqrt{x^2}=\lvert x\rvert\). So \(\lvert x\rvert=9\) gives \(x=9\) or \(x=-9\).

How would you solve \(\,x^2-81=0\,\) by factoring?

Factor: \((x-9)(x+9)=0\). Then \(x-9=0\) or \(x+9=0\), so \(x=9\) or \(x=-9\).

What are the solutions if \(\,x^2=0\,\) instead?

\(\sqrt{0}=0\), so \(x^2=0\) implies \(x=0\) only (no \(\pm\) difference).

Does \(\,x=\sqrt{81}\,\) always give all solutions?

No. \(\sqrt{81}=9\) gives only \(x=9\). The full set requires \(x=\pm 9\).

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