Q. \(x^2 = 144\)

Q: Solve x^2=144.

Take square roots: x=\pm\sqrt{144}=\pm 12. So x=12 or x=-12.

Q: What are all real solutions to x^2=144?

The real square roots are \pm 12. Therefore, the solutions are x=12 and x=-12.

Q: How do complex solutions look if x^2=144 had a negative right side?

For comparison, if x^2=-144, then x=\pm \sqrt{-144}=\pm 12i. (Here it’s real: x=\pm 12.)

Q: Why do we include both + and - when taking square roots?

Because squaring removes sign: (12)^2=(-12)^2=144. So both x=12 and x=-12 satisfy the equation.

Q: Can we write the solution as a set?

Yes: the solution set is \{ -12,\, 12 \}.

Q: What is the general method for equations like x^2=a?

Use x=\pm\sqrt{a} when a\ge 0. If a<0, solutions are complex: x=\pm i\sqrt{-a}.

Answer

We solve \(x^2=144\) by taking square roots of both sides:

\[
x=\pm\sqrt{144}=\pm 12
\]

Final result: \(x=12\) or \(x=-12\).

Detailed Explanation

We want to solve the equation \(x^2 = 144\).

Step 1: Take the square root of both sides

Since \(x^2\) is a perfect square, we take the square root of both sides:

\[
x^2 = 144
\]
\[
x = \pm \sqrt{144}
\]

Step 2: Simplify \(\sqrt{144}\)

\(144\) is a perfect square because \(12 \times 12 = 144\). So:

\[
\sqrt{144} = 12
\]

Step 3: Include both the positive and negative solutions

Because \(x^2\) equals \(144\), \(x\) can be either positive or negative:

\[
x = \pm 12
\]

Final Answer

The solutions are:

\[
x = 12 \quad \text{or} \quad x = -12
\]

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Algebra FAQ

Solve \(x^2=144\).

Take square roots: \(x=\pm\sqrt{144}=\pm 12\). So \(x=12\) or \(x=-12\).

What are all real solutions to \(x^2=144\)?

The real square roots are \(\pm 12\). Therefore, the solutions are \(x=12\) and \(x=-12\).

How do complex solutions look if \(x^2=144\) had a negative right side?

For comparison, if \(x^2=-144\), then \(x=\pm \sqrt{-144}=\pm 12i\). (Here it’s real: \(x=\pm 12\).)

Why do we include both \(+\) and \(-\) when taking square roots?

Because squaring removes sign: \((12)^2=(-12)^2=144\). So both \(x=12\) and \(x=-12\) satisfy the equation.

Can we write the solution as a set?

Yes: the solution set is \(\{ -12,\, 12 \}\).

What is the general method for equations like \(x^2=a\)?

Use \(x=\pm\sqrt{a}\) when \(a\ge 0\). If \(a<0\), solutions are complex: \(x=\pm i\sqrt{-a}\).

We solve x^2=144 step by step.
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