apply Pythagorean theorem then solve for x. (x+7)^2+(x)^2=(x+9)^2

To solve the given equation using the Pythagorean theorem:

\[
(x + 7)^2 + x^2 = (x + 9)^2
\]

Let's expand the squares:

1. Expand \((x + 7)^2\):
\[
(x + 7)^2 = x^2 + 14x + 49
\]

2. Expand \(x^2\):
\[
x^2 = x^2
\]

3. Expand \((x + 9)^2\):
\[
(x + 9)^2 = x^2 + 18x + 81
\]

Now substitute these into the equation:

\[
(x^2 + 14x + 49) + x^2 = x^2 + 18x + 81
\]

Combine like terms:

\[
2x^2 + 14x + 49 = x^2 + 18x + 81
\]

Next, move all terms to one side of the equation:

\[
2x^2 + 14x + 49 - x^2 - 18x - 81 = 0
\]

Simplify:

\[
x^2 - 4x - 32 = 0
\]

Now, solve this quadratic equation using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \(a = 1\), \(b = -4\), and \(c = -32\):

\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-32)}}{2(1)}
\]

Simplify further:

\[
x = \frac{4 \pm \sqrt{16 + 128}}{2}
\]
\[
x = \frac{4 \pm \sqrt{144}}{2}
\]
\[
x = \frac{4 \pm 12}{2}
\]

This gives two possible solutions:

\[
x = \frac{4 + 12}{2} = \frac{16}{2} = 8
\]

\[
x = \frac{4 - 12}{2} = \frac{-8}{2} = -4
\]

So, the solutions for \(x\) are:

\[
x = 8 \quad \text{or} \quad x = -4
\]

Would you like further details or have any questions?

Here are 8 related questions:

1. What is the Pythagorean theorem used for?
2. How do you determine the correct sign in the quadratic formula?
3. How do you verify if a solution satisfies the original equation?
4. What are the properties of quadratic equations?
5. Can you graph the equation \(x^2 - 4x - 32 = 0\)?
6. What is the discriminant in the quadratic formula, and what does it signify?
7. How do you complete the square to solve a quadratic equation?
8. What are the applications of the Pythagorean theorem in real life?

**Tip:** Always check both potential solutions when solving quadratic equations to ensure they satisfy the original equation.

Learn how to solve the equation (x+7)^2 + x^2 = (x+9)^2 step-by-step with the Pythagorean Theorem. This problem involves quadratic equations and Pythagorean theorem, suitable for high school students.

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