# 10.5: Square Root Applications

**At Grade**Created by: CK-12

**Practice**Square Root Applications

### Square Root Applications

We can use the methods we’ve learned so far in this section to find approximate solutions to quadratic equations, when taking the square root doesn’t give an exact answer.

#### Solving Quadratic Equations

1. Solve the following quadratic equations.

a) \begin{align*}x^2 - 3 = 0\end{align*}\begin{align*}\text{Isolate the} \ x^2: \qquad \qquad \qquad \qquad \qquad x^2 = 3\!\\ \text{Take the square root of both sides}: \qquad x = \sqrt{3} \ \text{and} \ x = -\sqrt{3}\end{align*}

**Answer:** \begin{align*}x \approx 1.73\end{align*} and \begin{align*}x \approx - 1.73\end{align*}

b) \begin{align*} 2x^2 - 9 = 0\end{align*}\begin{align*}\text{Isolate the} \ x^2: \qquad \qquad \qquad \qquad \qquad 2x^2 = 9 \ \text{so} \ x^2 = \frac{9} {2}\!\\ \text{Take the square root of both sides}: \qquad x = \sqrt{\frac{9} {2}} \ \text{and} \ x = -\sqrt{\frac{9} {2}}\end{align*}

**Answer:** \begin{align*}x \approx 2.12\end{align*} and \begin{align*}x \approx - 2.12\end{align*}

2. Solve the following quadratic equations.

a) \begin{align*}(2x + 5)^2 = 10\end{align*}

\begin{align*}\text{Take the square root of both sides}: & & 2x + 5 & = \sqrt{10} \ \text{and} \ 2x + 5 = -\sqrt{10}\\ \text{Solve both equations}: & & x & = \frac{-5 + \sqrt{10}} {2} \ \text{and} \ x = \frac{-5 -\sqrt{10}} {2}\end{align*}

**Answer:** \begin{align*}x \approx -0.92\end{align*} and \begin{align*}x \approx -4.08\end{align*}

b) \begin{align*}x^2 - 2x + 1 = 5\end{align*}\begin{align*}\text{Factor the right-hand-side}: & & (x - 1)^2 & = 5\\ \text{Take the square root of both sides}: & & x - 1 & = \sqrt{5} \ \text{and} \ x - 1 = -\sqrt{5}\\ \text{Solve each equation}: & & x & = 1 + \sqrt{5} \ \text{and} \ x = 1 - \sqrt{5}\end{align*}

**Answer:** \begin{align*}x \approx 3.24\end{align*} and \begin{align*}x \approx -1.24\end{align*}

**Solve Applications Using Quadratic Functions and Square Roots**

Quadratic equations are needed to solve many real-world problems. In this section, we’ll examine problems about objects falling under the influence of gravity. When objects are **dropped** from a height, they have no initial velocity; the force that makes them move towards the ground is due to gravity. The acceleration of gravity on earth is given by the equation

\begin{align*}g = -9.8 \ m/s^2 \quad \text{or} \quad g = -32 \ ft/s^2\end{align*}

The negative sign indicates a downward direction. We can assume that gravity is constant for the problems we’ll be examining, because we will be staying close to the surface of the earth. The acceleration of gravity decreases as an object moves very far from the earth. It is also different on other celestial bodies such as the moon.

The equation that shows the height of an object in free fall is

\begin{align*}y = \frac{1}{2}gt^2 + y_0\end{align*}

The term \begin{align*}y_0\end{align*} represents the initial height of the object, \begin{align*}t\end{align*} is time, and \begin{align*}g\end{align*} is the constant representing the force of gravity. You then plug in one of the two values for \begin{align*}g\end{align*} above, depending on whether you want the answer in feet or meters. Thus the equation works out to \begin{align*}y = -4.9t^2 + y_0\end{align*} if you want the height in meters, and \begin{align*}y = -16t^2 + y_0\end{align*} if you want it in feet.

#### Real-World Application: Object Falling

1. How long does it take a ball to fall from a roof to the ground 25 feet below?

\begin{align*}\text{Since we are given the height in feet, use equation}: & & y & = -16t^2 + y_0\\ \text{The initial height is} \ y_0 = 25 \ feet, \ \text{so}: & & y & = -16t^2 + 25\\ \text{The height when the ball hits the ground is} \ y = 0, \ \text{so}: & & 0 & = - 16t^2 + 25\\ \text{Solve for} \ t: & & 16t^2 & = 25\\ & & t^2 & = \frac{25} {16}\\ & & t & = \frac{5} {4} \ \text{or} \ t = - \frac{5} {4}\end{align*}

Since only positive time makes sense in this case, **it takes the ball 1.25 seconds to fall to the ground.**

2. A rock is dropped from the top of a cliff and strikes the ground 7.2 seconds later. How high is the cliff in meters?

\begin{align*}\text{Since we want the height in meters, use equation}: & & y & = -4.9t^2 + y_0\\ \text{The time of flight is} \ t = 7.2 \ seconds: & & y & = -4.9(7.2)^2 + y_0\\ \text{The height when the ball hits the ground is} \ y = 0, \ \text{so}: & & 0 & = -4.9 (7.2)^2 + y_0\\ \text{Simplify}: & & 0 & = -254 + y_0 \ \text{so} \ y_0 = 254\end{align*}

**The cliff is 254 meters high.**

### Example

#### Example 1

Victor throws an apple out of a window on the \begin{align*}10^{th}\end{align*} floor which is 120 feet above ground. One second later Juan throws an orange out of a \begin{align*}6^{th}\end{align*} floor window which is 72 feet above the ground. Which fruit reaches the ground first, and how much faster does it get there?

Let’s find the time of flight for each piece of fruit.

*Apple:*

\begin{align*}&\text{Since we have the height in feet, use this equation}: && y = -16t^2 + y_0\\ &\text{The initial height is} \ y_0 = 120 \ feet: && y = -16t^2 + 120\\ &\text{The height when the ball hits the ground is} \ y = 0, \ \text{so}: && 0 = -16t^2 + 120\\ &\text{Solve for} \ t: && 16t^2 = 120\\ &&& t^2 = \frac{120} {16} = 7.5\\ &&& \underline{t = 2.74} \ \text{or} \ t = -2.74 \ seconds\end{align*}

*Orange:*

\begin{align*}&\text{The initial height is} \ y_0 = 72 \ feet: & & 0 = -16t^2 + 72\\ &\text{Solve for} \ t: & & 16t^2 = 72\\ && & t^2 = \frac{72} {16} = 4.5\\ && & \underline{t = 2.12} \ \text{or} \ t = -2.12 \ seconds\end{align*}

The orange was thrown one second later, so add 1 second to the time of the orange: \begin{align*}t = 3.12 \ seconds\end{align*}

**The apple hits the ground first. It gets there 0.38 seconds faster than the orange.**

### Review

Solve the following quadratic equations.

- \begin{align*}x^2 = 11\end{align*}
- \begin{align*}5x^2 = 0.01\end{align*}
- \begin{align*}x^2 - 6 = 0\end{align*}
- \begin{align*}x^2 - 20 = 0\end{align*}
- \begin{align*}3x^2 + 14 = 0\end{align*}
- \begin{align*}(x - 6)^2 = 5\end{align*}
- \begin{align*}(x + 10)^2 = 2\end{align*}
- Susan drops her camera in the river from a bridge that is 400 feet high. How long is it before she hears the splash?
- It takes a rock 5.3 seconds to splash in the water when it is dropped from the top of a cliff. How high is the cliff in meters?
- Nisha drops a rock from the roof of a building 50 feet high. Ashaan drops a quarter from the top story window, 40 feet high, exactly half a second after Nisha drops the rock. Which hits the ground first?

### Review (Answers)

To view the Review answers, open this PDF file and look for section 10.5.

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
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Term | Definition |
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quadratic function |
A quadratic function is a function that can be written in the form , where , , and are real constants and . |

Square Root |
The square root of a term is a value that must be multiplied by itself to equal the specified term. The square root of 9 is 3, since 3 * 3 = 9. |

### Image Attributions

Here you'll learn how to approximate the solutions of quadratic equations involving square roots. You'll also learn how to solve applications using quadratic functions and square roots.

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