Q. \(x^2-49=0\)
Answer
We solve the quadratic equation \(x^2 – 49 = 0\).
Add \(49\) to both sides: \(x^2 = 49\).
Take the square root of both sides: \(x = \pm 7\).
Final result: \(x = 7\) or \(x = -7\).
Detailed Explanation
We want to solve the equation
\[
x^{2}-49=0
\]
Step 1: Move the constant term to the other side.
Start with
\[
x^{2}-49=0
\]
Add \(49\) to both sides so that the \(x^{2}\) term is alone.
\[
x^{2}-49+49=0+49
\]
The \(-49\) and \(+49\) cancel, leaving
\[
x^{2}=49
\]
Step 2: Take the square root of both sides.
Since \(x^{2}=49\), we have
\[
x=\pm\sqrt{49}
\]
Compute \(\sqrt{49}\).
\[
\sqrt{49}=7
\]
So the solutions are
\[
x=\pm 7
\]
Final Answer:
\[
x=7 \quad \text{or} \quad x=-7
\]
See full solution
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Algebra FAQ
How do I solve \(x^2-49=0\)?
Rewrite as \(x^2=49\), then take square roots: \(x=\pm 7\).
Can I use the difference of squares to factor \(x^2-49\)?
Yes. \(x^2-49=(x-7)(x+7)=0\) so \(x=7\) or \(x=-7\).
Why does \(x^2=49\) give two solutions?
Because squaring removes sign: both \(7^2\) and \((-7)^2\) equal \(49\).
What is the factoring method step-by-step?
Recognize \(49=7^2\). Then \(x^2-7^2=(x-7)(x+7)\). Set each factor to \(0\).
How do I verify the solutions in the original equation?
Substitute \(x=7\): \(49-49=0\). Substitute \(x=-7\): \(49-49=0\). Both check out.
What are the roots and their multiplicities?
The roots are \(x=7\) and \(x=-7\), each with multiplicity \(1\).
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