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

Answer:

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

The actual dimension is 18.28 by 3.28

Step-by-step 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:

$x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

Where

$a = 1$

$b = 15$

$c = -60$

So:

$x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-15 \± \sqrt{15^2 - 4*1*-60}}{2*1}$

$x = \frac{-15 \± \sqrt{225 + 240}}{2}$

$x = \frac{-15 \± \sqrt{465}}{2}$

$x = \frac{-15 \± \21.56}{2}$

Split:

$x = \frac{-15 + 21.56}{2}$ or $x = \frac{-15 - 21.56}{2}$

$x = \frac{6.56}{2}$ or $x = \frac{-36.36}{2}$

$x = 3.28$ 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