Q. \(x^{2}-81=0\)
Answer
We solve the equation by factoring:
\[
x^2 – 81 = 0
\]
\[
(x-9)(x+9)=0
\]
So the solutions are:
\[
x=9 \quad \text{or} \quad x=-9
\]
Final Result: \(x=9\) or \(x=-9\).
Detailed Explanation
We want to solve the equation
\[
x^2 – 81 = 0
\]
Step 1: Move the constant term to the other side. The goal is to isolate the \(x^2\) term.
\[
x^2 – 81 = 0
\]
Add \(81\) to both sides:
\[
x^2 – 81 + 81 = 0 + 81
\]
\[
x^2 = 81
\]
Step 2: Take the square root of both sides. Since \(x^2\) equals \(81\), \(x\) can be both the positive and negative square roots.
\[
x = \pm \sqrt{81}
\]
\[
\sqrt{81} = 9
\]
So:
\[
x = \pm 9
\]
Step 3: Write the two solutions.
\[
x = 9 \quad \text{or} \quad x = -9
\]
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Algebra FAQ
How do you solve \(x^2-81=0\)?
Rewrite as \(x^2=81\), so \(x=\pm 9\).
How do you factor \(x^2-81\)?
Use difference of squares: \(x^2-81=(x-9)(x+9)\).
Why is \(81\) written as a square in this problem?
Because \(81=9^2\), and \(x^2-a^2=(x-a)(x+a)\).
What is the solution using square root steps?
From \(x^2=81\), take square roots: \(x=\sqrt{81}=9\) or \(x=-\sqrt{81}=-9\).
How would you find all real solutions?
Solve \(x^2=81\), giving exactly two real solutions: \(x=9\) and \(x=-9\).
What are the roots of the polynomial \(x^2-81\)?
The roots are where the factors are zero: \(x-9=0\) and \(x+9=0\), so \(x=9\) and \(x=-9\).
How do you check your answers quickly?
Substitute: for \(x=9\), \(9^2-81=81-81=0\). For \(x=-9\), \((-9)^2-81=81-81=0\).
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