The side of a square has the length (x-2), while a rectangle has a length of (x-3) and a width of (x+4). If the area of the rectangle is twice the area of the square, what is the sum of the possible values of x? I CAN'T ACCESS A FILE SO PLEASE JUST LOOK AT THE PICTURE YOURSELF, AND THEN TELL ME. THANKS!

Sum = 4 + 5 = 9

Step-by-step explanation:

Equation

2*(x - 2)(x-2) = (x - 3)(x + 4)

Solution

2*(x^2 - 4x + 4) = x^2 + 4x - 3x - 12

2x^2 - 8x + 8  = x^2 +x - 12                     Subtract the right side from the left

x^2 - 9x + 20 = 0

(x - 4)(x - 5) = 0

The possible values for x are

x - 4 = 0

x = 4

x - 5 = 0

x = 5

Check

(4 - 2)^2 = (4) = 4

(x - 3)(x + 4) = (4 - 3)(8) = 8

This checks

(5 - 2)^2 = 3^2 = 9

(5 - 3)(5 + 4) = 2 * 9 = 18

.

step-by-step explanation:

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°