Q. \(x^2=16\)

Q: . Solve x^2=16.

. x=\pm 4. Since squaring removes sign, both positive and negative roots satisfy the equation.

Q: . What are the real solutions to x^2=16?

. The real solutions are x=4 and x=-4.

Q: . How do I solve x^2=16 by square roots?

. Take square roots: x=\sqrt{16} or x=-\sqrt{16}, giving x=\pm 4.

Q: . Are there complex solutions to x^2=16?

. Yes, but same values appear: x=\pm 4 are the complete solutions since 16>0.

Q: . Why do we need both x=4 and x=-4?

. Because x^2 equals the same value for x and -x. Squaring makes the result nonnegative.

Q: . What is the general solution to x^2=a when a=16?

. For a>0, x=\pm \sqrt{a}. Here x=\pm \sqrt{16}=\pm 4.

Q: . Check the solutions in x^2=16.

. 4^2=16 and (-4)^2=16, so both solutions are correct.

Answer

We solve \(x^2=16\) by taking square roots:

\[
x=\pm \sqrt{16}=\pm 4
\]

Final result: \(x=4\) or \(x=-4\).

Detailed Explanation

We want to solve the equation

\[
x^2 = 16
\]

Step 1: Take the square root of both sides.

Since both sides are perfect squares, we can use the square root rule. Remember: if \(x\) is squared, then \(x\) could be positive or negative.

\[
\sqrt{x^2} = \sqrt{16}
\]

This gives

\[
x = \pm \sqrt{16}
\]

Step 2: Simplify the square root.

\[
\sqrt{16} = 4
\]

So we have

\[
x = \pm 4
\]

Step 3: Write both solutions.

Therefore, the two values of \(x\) are

\[
x = 4 \quad \text{or} \quad x = -4
\]

Final answer:

\[
\boxed{x = 4 \text{ or } x = -4}
\]

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Algebra FAQ

. Solve \(x^2=16\).

. \(x=\pm 4\). Since squaring removes sign, both positive and negative roots satisfy the equation.

. What are the real solutions to \(x^2=16\)?

. The real solutions are \(x=4\) and \(x=-4\).

. How do I solve \(x^2=16\) by square roots?

. Take square roots: \(x=\sqrt{16}\) or \(x=-\sqrt{16}\), giving \(x=\pm 4\).

. Are there complex solutions to \(x^2=16\)?

. Yes, but same values appear: \(x=\pm 4\) are the complete solutions since \(16>0\).

. Why do we need both \(x=4\) and \(x=-4\)?

. Because \(x^2\) equals the same value for \(x\) and \(-x\). Squaring makes the result nonnegative.

. What is the general solution to \(x^2=a\) when \(a=16\)?

. For \(a>0\), \(x=\pm \sqrt{a}\). Here \(x=\pm \sqrt{16}=\pm 4\).

. Check the solutions in \(x^2=16\).

. \(4^2=16\) and \((-4)^2=16\), so both solutions are correct.

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