Question

Which of the following are solutions to 2x^2 - 8x-90? Select all that apply.

A. 2

B. -5

C. -7

D. 9

E. 15

F. 5

G. 7

Answer 1

The area of a rectangle is 56 cm. The length is 2 cm more than x and the width is 5 cm less than
twice x. Solve for x. Round to the nearest whole number.

x=

Answer 2

To find the solutions to the quadratic equation 2x^2 - 8x - 90, we first need to factor the equation:

2x^2 - 8x - 90 = 2(x^2 - 4x - 45) = 2(x - 9)(x + 5)

Setting each factor to zero to find the solutions:

x - 9 = 0 --> x = 9
x + 5 = 0 --> x = -5

The solutions are:

A. 2 (Not a solution)
B. -5 (Solution)
C. -7 (Not a solution)
D. 9 (Solution)
E. 15 (Not a solution)
F. 5 (Not a solution)
G. 7 (Not a solution)

Therefore, the solutions to 2x^2 - 8x - 90 are -5 and 9.

The correct options are B and D.

Answer 3

To solve for x, we first need to set up an equation based on the information given.

Let x be the width of the rectangle.

Given:
Area of the rectangle = 56 cm^2
Length = x + 2
Width = 2x - 5

The formula for the area of a rectangle is length * width:

(x + 2)*(2x - 5) = 56

Expand the left side of the equation:

2x^2 - 5x + 4x - 10 = 56
2x^2 - x - 10 = 56
2x^2 - x - 66 = 0

Now we have a quadratic equation in the form of Ax^2 + Bx + C = 0. We can solve this by factoring or using the quadratic formula.

Factoring is not straightforward, so let's use the quadratic formula:

x = [-(-1) ± sqrt((-1)^2 - 4*2*(-66))]/(2*2)
x = [1 ± sqrt(1 + 528)]/4
x = [1 ± sqrt(529)]/4
x = [1 ± 23]/4

This gives us two solutions:
x = (1 + 23)/4 = 24/4 = 6
x = (1 - 23)/4 = -22/4 = -5.5

Since x should be a positive value (width cannot be negative), x is approximately 6 when rounded to the nearest whole number.

Therefore, x = 6.