Question

A piece of cardboard has the dimensions (x + 15) inches by (x) inches with the area of 60 in 2 . Write the quadratic equation that represents this and show your work. Then find the possible value(s) for x, and find the actual dimensions of the postcard.

Answer 1

`x^2 + 15x - 60 = 0`

The actual dimension is 18.28 by 3.28

Explanation:

Given

Dimension:

`(x + 15)\ by\ x`

`Area = 60in^2`

Required

Determine the quadratic equation and get the possible values of x

Solving (a): Quadratic Equation.

The cardboard is rectangular in shape.

Hence, Area is calculated as thus:

`Area = Length * Width`

`60= (x + 15) * x`

Open Bracket

`60= x^2 + 15x`

Subtract 60 from both sides

`x^2 + 15x - 60 = 0`

Hence, the above represents the quadratic equation

Solving (b): The actual dimension

First, we need to solve for x

This can be solved using quadratic formula:

`x = (-b \± √(b^2 - 4ac))/(2a)`

Where

`a = 1`

`b = 15`

`c = -60`

So:

`x = (-b \± √(b^2 - 4ac))/(2a)`

`x = (-15 \± √(15^2 - 4*1*-60))/(2*1)`

`x = (-15 \± √(225 + 240))/(2)`

`x = (-15 \± √(465))/(2)`

`x = (-15 \± \21.56)/(2)`

Split:

`x = (-15 + 21.56)/(2)` or
`x = (-15 - 21.56)/(2)`

`x = (6.56)/(2)` or
`x = (-36.36)/(2)`

`x = 3.28` or
`x = -18.18`

But length can't be negative;

So:

`x = 3.28`

The actual dimensions:
`(x + 15)\ by\ x` is

`Length =3.28 +15`

`Length =18.28`

`Width = x`

`Width =3.28`

The actual dimension is 18.28 by 3.28