Q. \(x^2 = 7\).

Q: What are the solutions to x^2=7?

Rewrite as x=\pm \sqrt{7}. So the solutions are x=\sqrt{7} and x=-\sqrt{7}.

Q: How do you solve x^2=7 step by step?

Take square roots: x^2=7 \Rightarrow x=\pm \sqrt{7}. Because squaring loses the sign, both positive and negative roots are valid.

Q: Why do we use \pm when solving x^2=7?

Squaring makes x and -x produce the same result. Since (\sqrt{7})^2=7 and (-\sqrt{7})^2=7, both signs must be included.

Q: Is x=\sqrt{7} the only solution?

No. x=\sqrt{7} is one solution, and x=-\sqrt{7} is the other, since both square to 7.

Q: What is an approximate decimal value for \sqrt{7}?

\sqrt{7}\approx 2.64575. So x\approx 2.64575 or x\approx -2.64575.

Answer

We solve \(x^2=7\) by taking square roots of both sides.

\[
x=\sqrt{7}\ \text{or}\ x=-\sqrt{7}
\]

Detailed Explanation

We are asked to solve the equation

\[
x^2 = 7
\]

Step 1: Take the square root of both sides.

Because \(x^2\) is a perfect square, we take the square root of both sides. Remember that square roots have two possible values: one positive and one negative.

\[
\sqrt{x^2} = \sqrt{7}
\]

So we get:

\[
x = \pm \sqrt{7}
\]

Step 2: State the solutions.

The equation \(x^2 = 7\) has two real solutions:

\[
x = \sqrt{7}
\]

and

\[
x = -\sqrt{7}
\]

Final Answer:

\[
x = \pm \sqrt{7}
\]

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Algebra FAQ

What are the solutions to \(x^2=7\)?

Rewrite as \(x=\pm \sqrt{7}\). So the solutions are \(x=\sqrt{7}\) and \(x=-\sqrt{7}\).

How do you solve \(x^2=7\) step by step?

Take square roots: \(x^2=7 \Rightarrow x=\pm \sqrt{7}\). Because squaring loses the sign, both positive and negative roots are valid.

Why do we use \(\pm\) when solving \(x^2=7\)?

Squaring makes \(x\) and \(-x\) produce the same result. Since \((\sqrt{7})^2=7\) and \((-\sqrt{7})^2=7\), both signs must be included.

Is \(x=\sqrt{7}\) the only solution?

No. \(x=\sqrt{7}\) is one solution, and \(x=-\sqrt{7}\) is the other, since both square to \(7\).

What is an approximate decimal value for \(\sqrt{7}\)?

\(\sqrt{7}\approx 2.64575\). So \(x\approx 2.64575\) or \(x\approx -2.64575\).

What happens if the equation is \(x^2=-7\) instead?

There are no real solutions because a real number squared cannot be negative. Complex solutions exist: \(x=\pm i\sqrt{7}\).

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