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

554.4

step-by-step explanation:

44*12.6=544.4

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15