Problem 84 - Algebra for College Students

Question

Find the exact solution \((s)\) to each problem. If the solution(s) are irrational, then also find approximate solution(s) to the nearest tenth. Diagonal of a square. The diagonal of a square is \(2 \mathrm{m}\) longer than a side. Find the length of a side.

Step-by-Step Solution

Answer

The exact solution is \( 2 + 2 \sqrt{2} \) meters, which approximately equals 4.8 meters.

Step 1 - Define Variables

Let the length of the side of the square be denoted as \(s\). The diagonal will then be \(s + 2\) meters, since it is 2 meters longer than a side.

Step 2 - Apply the Pythagorean Theorem

In a square, the relationship between the side length and the diagonal is given by the Pythagorean theorem: \( \text{diagonal}^2 = \text{side}^2 + \text{side}^2 \). Substituting the given values, we get \( (s + 2)^2 = s^2 + s^2 \).

Step 3 - Simplify the Equation

Simplify the equation: \[ (s + 2)^2 = 2s^2 \] \[ s^2 + 4s + 4 = 2s^2 \]

Step 4 - Rearrange the Equation

Rearrange the equation to form a standard quadratic equation: \[ s^2 - 4s - 4 = 0 \]

Step 5 - Use the Quadratic Formula

Solve for \(s\) using the quadratic formula \( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = -2 \), and \( c = -4 \).

Step 6 - Calculate the Discriminant

Calculate the discriminant: \[ b^2 - 4ac = (-4)^2 - 4(1)(-4) = 16 + 16 = 32 \]

Step 7 - Find the Solution(s)

Substitute the discriminant back into the quadratic formula: \[ s = \frac{4 \pm \sqrt{32}}{2} = \frac{4 \pm 4\sqrt{2}}{2} = 2 \pm 2\sqrt{2} \]. This yields the exact solutions: \( s = 2 + 2\sqrt{2} \) and \( s = 2 - 2\sqrt{2} \). Since a length cannot be negative, the valid solution is \( s = 2 + 2 \sqrt{2} \).

Step 8 - Approximate the Solution

Approximate \( \sqrt{2} \approx 1.414 \), so \( s \approx 2 + 2(1.414) \approx 4.8 \).

Key Concepts

Quadratic Equations

To solve the problem, we need to understand quadratic equations. These are equations of the form \[ ax^2 + bx + c = 0 \].They are critical in many areas of math because they allow us to solve for unknown variables.In our exercise, the quadratic equation is derived from the properties of a square and its diagonal:
\[ s^2 - 4s - 4 = 0 \].
Here, - \(a\) is the coefficient of the squared term - \(b\) is the coefficient of the linear term- \(c\) is the constant term.The general solution to a quadratic equation can be found using the quadratic formula.

Solving for Side Length

We derived the quadratic equation from the given conditions in the problem and now need to solve for the side length \(s\). By using the quadratic formula:
\[ s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \],we substitute the respective coefficients:

• \(a = 1\)
• \(b = -4\)
• \(c = -4\)

Plugging these values into the quadratic formula will help us find the possible values for \(s\).

Discriminant Calculations

The value inside the square root in the quadratic formula is called the discriminant, given by: \[ b^2 - 4ac \].
The discriminant gives important information about the nature of the roots of the quadratic equation:

• If the discriminant is positive, the quadratic equation has two distinct real solutions.
• If the discriminant is zero, there is exactly one real solution (a repeated root).
• If the discriminant is negative, the quadratic equation has two complex solutions.

For our exercise:
\[ ( - 4 )^2 - 4 (1) (-4) = 16 + 16 = 32 \].
Since the discriminant (32) is positive, we know there are two real solutions for the side length \(s\).

Approximation of Irrational Numbers

After solving the quadratic formula, the solutions might include irrational numbers (numbers that cannot be expressed as a simple fraction). In this problem, the exact solution given is:
\[ s = 2 + 2\sqrt{2} \].
Since this involves \( \sqrt{2} \), which is an irrational number, we approximate its value. Roughly,\( \sqrt{2} \approx 1.414 \).
So, we simplify our solution:
\[ s \approx 2 + 2(1.414) \approx 2 + 2.828 \approx 4.8 \].
This approximation helps us get a practical understanding of the side length.
Approximations are crucial, especially when exact values are difficult to interpret or work with in real life.