To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the equal sign and the answer on the other.

2x + 4 = \( \frac{x}{8} \)
8 x (2x + 4) = x
(8 x 2x) + (8 x 4) = x
16x + 32 = x
16x + 32 - x = 0
16x - x = -32
15x = -32
x = \( \frac{-32}{15} \)
x = -2\(\frac{2}{15}\)

To factor a quadratic expression, apply the FOIL method (First, Outside, Inside, Last) in reverse. First, find the two Last terms that will multiply to produce -1 as well and sum (Inside, Outside) to equal 0. For this problem, those two numbers are -1 and 1. Then, plug these into a set of binomials using the square root of the First variable (y²):

y² - 1
y² + (-1 + 1)y + (-1 x 1)
(y - 1)(y + 1)

A monomial contains one term, a binomial contains two terms, and a polynomial contains more than two terms.

To find the diagonal we need to know the length of one of the square's sides. We know the area and the area of a square is the length of one side squared:

a = s²

so the length of one side of the square is:

s = \( \sqrt{a} \) = \( \sqrt{25} \) = 5

The Pythagorean theorem defines the square of the hypotenuse (diagonal) of a triangle with a right angle as the sum of the squares of the other two sides:

c² = a² + b²
c² = 5² + 5²
c² = 50
c = \( \sqrt{50} \) = \( \sqrt{25 x 2} \) = \( \sqrt{25} \) \( \sqrt{2} \)
c = 5\( \sqrt{2} \)

The Pythagorean theorem defines the relationship between the side lengths of a right triangle. The length of the hypotenuse squared (c²) is equal to the sum of the two perpendicular sides squared (a² + b²): c² = a² + b² or, solved for c, \(c = \sqrt{a + b}\)