Q. \(x^2=25\)
Answer
We solve \(x^2=25\) by taking square roots: \(\,x=\pm 5\).
\[
x=\pm 5
\]
Detailed Explanation
We want to solve the equation
\[
x^2 = 25
\]
Step 1: Take the square root of both sides.
Since \(x^2\) is a square, \(x\) can be positive or negative. So we write:
\[
\sqrt{x^2} = \sqrt{25}
\]
This gives:
\[
x = \pm 5
\]
Step 2: State both solutions.
So the two values of \(x\) that satisfy the equation are:
\[
x = 5 \quad \text{or} \quad x = -5
\]
Final Answer:
\[
x \in \{5,\,-5\}
\]
See full solution
Algebra FAQ
How do I solve \(x^2=25\)?
Take square roots: \(x=\pm\sqrt{25}=\pm 5\).
Why are there two solutions?
Because squaring removes signs: both \(5^2\) and \((-5)^2\) equal \(25\). So \(x=5\) or \(x=-5\).
What is the step when taking square roots of both sides?
From \(x^2=25\), write \(x=\pm\sqrt{25}\) since \(\sqrt{x^2}=|x|\), then \( |x|=5 \).
What if the equation is \(x^2=16\)?
Similarly, \(x=\pm\sqrt{16}=\pm 4\). Always two real solutions when the right side is positive perfect squares.
How do I check my solutions?
Substitute: \(5^2=25\) and \((-5)^2=25\). Both satisfy \(x^2=25\).
What happens if \(x^2=-25\)?
No real solutions because \(x^2\ge 0\). Over complex numbers, \(x=\pm 5i\).
Check x^2=25 with steps.
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