Application Of Quadratic Equations To Real Life Situations.

Subsection 1.3.6 Application Of Quadratic Equations To Real Life Situations.

Quadratic equations are a type of mathematical equation that can be used to describe many different real-life situations. Whether you’re throwing a ball in the air, trying to maximize the area of a garden, or calculating profits for a business, quadratic equations can help solve problems that involve relationships with squared terms.

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In this section, we will learn how to recognize real-life situations that can be modeled by quadratic equations, how to set them up, and how to solve them step by step.

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A quadratic equation is an equation that can be written in the form:

\begin{equation*} ax^2 + bx + c = 0 \end{equation*}

where,

\(a\text{,}\) \(b\) and \(c\) are real numbers or constants,
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\(x\) is the unknown variable
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and \(a \neq 0\) beacuse if \(a\) is \(0,\) it would npt be a quadratic
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Quadratic equations show up in many situations, especially when something is changing in a way that involves squares (like height, area, or profit).

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Lets take a look at some of the situations Quadratic Equations is involved in real life situations.

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\(1. \, Projectile \, Motion (Throwing \, a \, Ball).\)

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One of the most common real-life situations for quadratic equations is projectile motion, such as when you throw a ball in the air. The height of the ball over time can be described by a quadratic equation.

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Example 1.3.19.

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A rock is dropped from a height of \(50\) meters. Its height above the ground at time \(t\) is given by

\begin{equation*} h(t) = 5t^2 + 50. \end{equation*}

Use factorization to determine how long it will take for the rock to reach the ground.

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Solution.

The height of the rock above the ground is given by:

\begin{equation*} h(t) = -5t^2 + 50 \end{equation*}

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To find when the rock reaches the ground, set \(h(t) = 0\text{:}\)

\begin{equation*} -5t^2 + 50 = 0 \end{equation*}

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Simplify the equation:

\begin{equation*} -5t^2 = -50 \end{equation*}

\begin{equation*} t^2 = 10 \end{equation*}

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Solve for \(t\text{:}\)

\begin{equation*} t = \pm \sqrt{10} \end{equation*}

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Since time cannot be negative, \(t = \sqrt{10}\text{.}\)

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Therefore, the rock will take approximately \(\sqrt{10} \approx 3.16\) seconds to reach the ground.

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\(2. \, Maximizing \, Area \, (Optimization \, Problem).\)

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Quadratic equations are also used in optimization problems. These are problems where you want to maximize or minimize something, like the area of a garden.

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Example 1.3.20.

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The length of a rectangle is \(3\) meters more than its width. If the area of the rectangle is \(54\) square meters, find the dimensions of the rectangle by factoring the quadratic equation.

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Solution.

Let the width of the rectangle be \(x\text{.}\) Then the length is \(x + 3\text{.}\)

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The area of the rectangle is given by:

\begin{equation*} x(x + 3) = 54 \end{equation*}

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Expand and rearrange into standard quadratic form:

\begin{equation*} x^2 + 3x - 54 = 0 \end{equation*}

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Factorize the quadratic equation:

\begin{equation*} (x + 9)(x - 6) = 0 \end{equation*}

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Solve for \(x\text{:}\)

\begin{equation*} x = -9 \, \text{or} \, x = 6 \end{equation*}

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Since width cannot be negative, \(x = 6\text{.}\)

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The dimensions of the rectangle are:

\begin{equation*} \text{Width} = 6 \, \text{meters}, \, \text{Length} = 6 + 3 = 9 \, \text{meters}. \end{equation*}

Since the width cannot be negative, the width is \(6\) meters, and lenght is \(6 + 3 = 9\) meters.

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\(3. Business \, Applications (Profit \, Function)\)

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Businesses often use quadratic equations to model their profit. For example, the profit a company makes from selling a product can be modeled by a quadratic equation.

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Example 1.3.21.

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A company’s profit \(P(x)\) from selling \(x\) units of a product is given by the equation:

\begin{equation*} P(x) = -x^2 + 15x - 50 \end{equation*}

Find how many units the company needs to sell to have no profit (i.e., when the profit is zero).

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Solution.

To find when the company has no profit, set \(P(x) = 0\text{:}\)

\begin{equation*} -x^2 + 20x - 50 = 0 \end{equation*}

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Multiply through by \(-1\) to simplify:

\begin{equation*} x^2 - 20x + 50 = 0 \end{equation*}

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Factorize the quadratic equation:

\begin{equation*} (x - 10)(x - 5) = 0 \end{equation*}

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Solve for \(x\text{:}\)

\begin{equation*} x = 10 or x = 5 \end{equation*}

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Therefore, the company will have no profit when it sells either \(5\) units or \(10\) units.

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Exercises Exercises

1.

A stone is thrown into the air from a height of \(4\) meters with an initial velocity of \(8\) meters per second. The height of the stone at time \(t\) is given by:

\begin{equation*} h(t) = -5t^2 + 8t + 4 \end{equation*}

Find when the stone reaches the ground.

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2.

A farmer has \(200\) meters of fencing. He wants to build a rectangular garden. The length is \(50\) meters longer than the width. What should the dimensions be to maximize the area?

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3.

A school’s profit function is given by:

\begin{equation*} P(x) = -x^2 + 30x - 100 \end{equation*}

Find the number of units the company must sell to achieve zero profit.

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4.

A water fountain shoots water into the air, and its height at any time \(t\) (in seconds) is given by the equation:

\begin{equation*} h(t) = -4.9t^2 + 15t + 2 \end{equation*}

Find the time it will take for the water to return to the ground (i.e., when \(h(t)=0\)).

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5.

Solve the following quadratic equation:

\begin{equation*} x^2 - 7x + 12 = 0 \end{equation*}

Find the value of \(x.\)

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6.

A company finds that the revenue \(R(x)\) it generates from selling \(x\) units of a product is given by the quadratic equation:

\begin{equation*} R(x) = 2x^2 + 40x \end{equation*}

Find the number of units \(x\) that the company needs to sell to maximize its revenue.

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7.

The path of a car is represented by the quadratic equation

\begin{equation*} y = 2x^2 - 8x, \end{equation*}

where \(x\) represents time in seconds and \(y\) represents the car’s position. Find the time when the car reaches its starting position by factorizing the equation.

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