Miriam reduced a square photo by cutting 3 inches away from the length and the width so it will fit in her photo album. The area of the reduced photo is 64 square inches. In the equation (x – 3)2 = 64, x represents the side measure of the original photo.

What were the dimensions of the original photo?

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Professor1994 Professor1994 · Answer 1 · 2018-11-11T08:11:01+01:00

Answer: Solving the quadratic equation, the dimensions of the original photo are 11 inches by 11 inches.

First option 11 inches by 11 inches

Solution:

Side measure of the original photo: x

(x-3)^2=64

Solving for x: Square root both sides of the equation:

√[(x-3)^2]=±√64

x-3=±8

Adding 3 both sides of the equation:

x-3+3=±8+3

x=±8+3

x1=-8+3→x1=-5<0 (negative), it's not a solution because the side measure of the original photo can be a negative number.

x2=8+3→x2=11>0 (positive). Ok

The solution is x=11 inches

psm22415 psm22415 · Answer 2 · 2022-01-25T02:20:17+01:00

The dimension of the original photo is 11 inches by 11 inches.

The area of the reduced photo is 64 square inches.

In the equation (x – 3)2 = 64, x represents the side measure of the original photo.

What were the dimensions of the original photo?

The area of the reduced photo is 64 square inches.

In the equation (x – 3)2 = 64, x represents the side measure of the original photo.

In the equation, x represents the side measure of the original photo.

Therefore,

Solving the equation for the value of x,

[tex]\rm (x-3)^2=64\\ \\ (x-3)^2=8^2\\ \\ x-3=\pm 8\\ \\ x-3 = 8, \ x=8+3, \ x=11\\ \\ x-3=-8, \ x=-8+3, \ x = -5[/tex]

The length and width can not be negative.

Therefore, the dimension of the original photo is 11 inches by 11 inches.

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