Q. \(x^2 – 36 = 0\)

Q: How do you solve x^2-36=0 by factoring?

Rewrite as x^2-6^2=0. Then (x-6)(x+6)=0. So x=6 or x=-6.

Q: What are the solutions using the square root method?

x^2=36. Then x=\pm\sqrt{36}=\pm 6. So x=6 and x=-6.

Q: How do you solve using the quadratic formula?

For x^2+0x-36=0, a=1,b=0,c=-36. Then x=\frac{-0\pm\sqrt{0-4(1)(-36)}}{2}=\pm 6.

Q: Why does x^2=36 give two solutions instead of one?

Squaring loses sign information because (6)^2=(-6)^2=36. So both x=6 and x=-6 satisfy the original equation.

Q: What is the discriminant, and what does it tell you?

Discriminant D=b^2-4ac=0-4(1)(-36)=144. Since D>0, there are two distinct real solutions: x=\pm 6.

Q: How can you check your solutions in the original equation?

Substitute x=6: 6^2-36=36-36=0. Substitute x=-6: (-6)^2-36=36-36=0. Both work.

Answer

We solve \(x^2-36=0\) by moving \(36\) to the other side:

\[
x^2=36
\]

Take square roots:

\[
x=\pm 6
\]

Final result: \(x=6\) or \(x=-6\).

Detailed Explanation

We want to solve the equation:

\[
x^2 – 36 = 0
\]

Step 1: Move the constant term to the other side.

Add \(36\) to both sides so that the \(x^2\) term is by itself.

\[
x^2 – 36 + 36 = 0 + 36
\]

\[
x^2 = 36
\]

Step 2: Take the square root of both sides.

When you take a square root, you must include both the positive and negative solutions because \(x^2\) is the same for \(x\) and \(-x\).

\[
x = \pm \sqrt{36}
\]

Step 3: Simplify the square root.

Since \(36 = 6^2\), we have \(\sqrt{36} = 6\).

\[
x = \pm 6
\]

Final answer:

\[
x = 6 \quad \text{or} \quad x = -6
\]

See full solution

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Algebra FAQ

How do you solve \(x^2-36=0\) by factoring?

Rewrite as \(x^2-6^2=0\). Then \((x-6)(x+6)=0\). So \(x=6\) or \(x=-6\).

What are the solutions using the square root method?

\(x^2=36\). Then \(x=\pm\sqrt{36}=\pm 6\). So \(x=6\) and \(x=-6\).

How do you solve using the quadratic formula?

For \(x^2+0x-36=0\), \(a=1,b=0,c=-36\). Then \(x=\frac{-0\pm\sqrt{0-4(1)(-36)}}{2}=\pm 6\).

Why does \(x^2=36\) give two solutions instead of one?

Squaring loses sign information because \((6)^2=(-6)^2=36\). So both \(x=6\) and \(x=-6\) satisfy the original equation.

What is the discriminant, and what does it tell you?

Discriminant \(D=b^2-4ac=0-4(1)(-36)=144\). Since \(D>0\), there are two distinct real solutions: \(x=\pm 6\).

How can you check your solutions in the original equation?

Substitute \(x=6\): \(6^2-36=36-36=0\). Substitute \(x=-6\): \((-6)^2-36=36-36=0\). Both work.

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