The first step to solve a quadratic expression that's not set to zero is to solve the equation so that it is set to zero:

b² + 5b + 1 = 3b + 4
b² + 5b + 1 - 4 = 3b
b² + 5b - 3b - 3 = 0
b² + 2b - 3 = 0

Next, factor the quadratic equation:

b² + 2b - 3 = 0
(b - 1)(b + 3) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (b - 1) or (b + 3) must equal zero:

If (b - 1) = 0, b must equal 1
If (b + 3) = 0, b must equal -3

So the solution is that b = 1 or -3

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 = \( \sqrt{a} \) = \( \sqrt{81} \) = 9

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 = 9² + 9²
c² = 162
c = \( \sqrt{162} \) = \( \sqrt{81 x 2} \) = \( \sqrt{81} \) \( \sqrt{2} \)
c = 9\( \sqrt{2} \)

The perimeter of a triangle is the sum of the lengths of its sides:

p = x + y + z
p = 12cm + 6cm + 6cm = 24cm

To multiply monomials, multiply the coefficients (the numbers that come before the variables) of each term, add the exponents of like variables, and multiply the different variables together.

(2a)(3ab) + (7a²)(3b)
(2 x 3)(a x a x b) + (7 x 3)(a² x b)
(6)(a¹⁺¹ x b) + (21)(a²b)
6a²b + 21a²b
27a²b

The slope of this line is the change in y divided by the change in x. The endpoints of this line segment are at (-2, -7) and (2, -1) so the slope becomes: