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# 10.3: Quadratic Equations by Square Roots

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## Learning Objectives

• Solve quadratic equations involving perfect squares.
• Approximate solutions of quadratic equations.
• Solve real-world problems using quadratic functions and square roots.

## Introduction

So far you know how to solve quadratic equations by factoring. However, this method works only if a quadratic polynomial can be factored. In the real world, most quadratics can’t be factored, so now we’ll start to learn other methods we can use to solve them. In this lesson, we’ll examine equations in which we can take the square root of both sides of the equation in order to arrive at the result.

## Solve Quadratic Equations Involving Perfect Squares

Let’s first examine quadratic equations of the type

$x^2 - c = 0$

We can solve this equation by isolating the $x^2$ term: $x^2 = c$

Once the $x^2$ term is isolated we can take the square root of both sides of the equation. Remember that when we take the square root we get two answers: the positive square root and the negative square root:

$x = \sqrt{c} \qquad \text{and} \qquad x = -\sqrt{c}$

Often this is written as $x = \pm \sqrt{c}$.

Example 1

a) $x^2 - 4 = 0$

b) $x^2 - 25 = 0$

Solution

a) $x^2 - 4 = 0$

Isolate the $x^2$: $x^2 = 4$

Take the square root of both sides: $x = \sqrt{4}$ and $x = - \sqrt{4}$

The solutions are $x = 2$ and $x = -2$.

b) $x^2 - 25 = 0$

Isolate the $x^2$: $x^2 = 25$

Take the square root of both sides: $x = \sqrt{25}$ and $x = - \sqrt{25}$

The solutions are $x = 5$ and $x = -5$.

We can also find the solution using the square root when the $x^2$ term is multiplied by a constant—in other words, when the equation takes the form

$ax^2 - c = 0$

We just have to isolate the $x^2$:

$ax^2 & = b\\x^2 & = \frac{b}{a}$

Then we can take the square root of both sides of the equation:

$x = \sqrt{\frac{b}{a}} \qquad \text{and} \qquad x = - \sqrt{ \frac{b}{a}}$

Often this is written as: $x = \pm \sqrt{\frac{b}{a}}$.

Example 2

a) $9x^2 - 16 = 0$

b) $81x^2 - 1 = 0$

Solution

a) $9x^2 - 16 = 0$

Isolate the $x^2$:

$9x^2 & = 16\\x^2 & = \frac{16}{9}$

Take the square root of both sides: $x = \sqrt{\frac{16}{9}}$ and $x = - \sqrt{ \frac{16}{9}}$

Answer: $x = \frac{4}{3}$ and $x = - \frac{4}{3}$

b) $81x^2 - 1 = 0$

Isolate the $x^2$:

$81x^2 & = 1\\x^2 & = \frac{1}{81}$ Take the square root of both sides: $x = \sqrt{\frac{1}{81}}$ and $x = - \sqrt{ \frac{1}{81}}$

Answer: $x = \frac{1}{9}$ and $x = - \frac{1}{9}$

As you’ve seen previously, some quadratic equations have no real solutions.

Example 3

a) $x^2 + 1 = 0$

b) $4x^2 + 9 = 0$

Solution

a) $x^2 + 1 = 0$

Isolate the $x^2$: $x^2 = -1$

Take the square root of both sides: $x = \sqrt{-1}$ and $x = - \sqrt{-1}$

Square roots of negative numbers do not give real number results, so there are no real solutions to this equation.

b) $4x^2 + 9 = 0$

Isolate the $x^2$:

$4x^2 & = -9\\x^2 & = - \frac{9}{4}$ Take the square root of both sides: $x = \sqrt{ - \frac{9}{4}}$ and $x = - \sqrt{ - \frac{9}{4}}$

There are no real solutions.

We can also use the square root function in some quadratic equations where one side of the equation is a perfect square. This is true if an equation is of this form:

$(x - 2)^2 = 9$

Both sides of the equation are perfect squares. We take the square root of both sides and end up with two equations: $x - 2 = 3$ and $x - 2 = -3$.

Solving both equations gives us $x = 5$ and $x = -1$.

Example 4

a) $(x - 1)^2 = 4$

b) $(x + 3)^2 = 1$

Solution

a) $(x - 1)^2 = 4$

$\text{Take the square root of both sides}: & & x - 1 & = 2 \ \text{and} \ x - 1 = -2\\\text{Solve each equation}: & & x & = 3 \ \text{and} \ x = -1$

Answer: $x =3$ and $x = -1$

b) $(x + 3)^2 = 1$

$\text{Take the square root of both sides}: & & x + 3 & = 1 \ \text{and} \ x + 3 = -1\\\text{Solve each equation}: & & x & = -2 \ \text{and} \ x = -4$

Answer: $x = -2$ and $x = -4$

It might be necessary to factor the right-hand side of the equation as a perfect square before applying the method outlined above.

Example 5

a) $x^2 + 8x + 16 = 25$

b) $4x^2 - 40x + 25 = 9$

Solution

a) $x^2 + 8x + 16 = 25$

$& \text{Factor the right-hand-side}: & & x^2 + 8x + 16 = (x + 4)^2 \quad \text{so} \quad (x + 4)^2 = 25\\& \text{Take the square root of both sides}: & & x + 4 = 5 \ \text{and} \ x + 4 = -5 \\& \text{Solve each equation}: & & x = 1 \ \text{and} \ x = -9$

Answer: $x = 1$ and $x = -9$

b) $4x^2 - 20x + 25 = 9$

$& \text{Factor the right-hand-side}: & & 4x^2 - 20x + 25 = (2x - 5)^2 \quad \text{so} \quad (2x - 5)^2 = 9\\& \text{Take the square root of both sides}: & & 2x - 5 = 3 \ \text{and} \ 2x - 5 = -3 \\& \text{Solve each equation}: & & 2x = 8 \ \text{and} \ 2x = 2$

Answer: $x = 4$ and $x =1$

## Approximate Solutions of Quadratic Equations

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.

Example 6

a) $x^2 - 3 = 0$

b) $2x^2 - 9 = 0$

Solution

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

Answer: $x \approx 1.73$ and $x \approx - 1.73$

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

Answer: $x \approx 2.12$ and $x \approx - 2.12$

Example 7

a) $(2x + 5)^2 = 10$

b) $x^2 - 2x + 1 = 5$

Solution

a) $\text{Take the square root of both sides}: & & 2x + 5 & = \sqrt{10} \ \text{and} \ 2x + 5 = -\sqrt{10}\\\text{Solve both equations}: & & x & = \frac{-5 + \sqrt{10}} {2} \ \text{and} \ x = \frac{-5 -\sqrt{10}} {2}$

Answer: $x \approx -0.92$ and $x \approx -4.08$

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

Answer: $x \approx 3.24$ and $x \approx -1.24$

## Solve Real-World Problems 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

$g = -9.8 \ m/s^2 \quad \text{or} \quad g = -32 \ ft/s^2$

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

$y = \frac{1}{2}gt^2 + y_0$

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

Example 8

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

Solution

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

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

Example 9

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?

Solution

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

The cliff is 254 meters high.

Example 10

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

Solution

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

Apple:

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

Orange:

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

The orange was thrown one second later, so add 1 second to the time of the orange: $t = 3.12 \ seconds$

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

## Review Questions

1. $x^2 - 1 = 0$
2. $x^2 - 100 = 0$
3. $x^2 + 16 = 0$
4. $9x^2 - 1 = 0$
5. $4x^2 - 49 = 0$
6. $64x^2 - 9 = 0$
7. $x^2 - 81 = 0$
8. $25x^2 - 36 = 0$
9. $x^2 + 9 = 0$
10. $x^2 - 16 = 0$
11. $x^2 - 36 = 0$
12. $16x^2 - 49 = 0$
13. $(x - 2)^2 = 1$
14. $(x + 5)^2 = 16$
15. $(2x - 1)^2 - 4 = 0$
16. $(3x + 4)^2 = 9$
17. $(x - 3)^2 + 25 = 0$
18. $x^2 - 6 = 0$
19. $x^2 - 20 = 0$
20. $3x^2 + 14 = 0$
21. $(x - 6)^2 = 5$
22. $(4x + 1)^2 - 8 = 0$
23. $x^2 - 10x + 25 =9$
24. $x^2 + 18x + 81 = 1$
25. $4x^2 - 12x + 9 = 16$
26. $(x + 10)^2 = 2$
27. $x^2 + 14x + 49 = 3$
28. $2(x + 3)^2 = 8$
29. Susan drops her camera in the river from a bridge that is 400 feet high. How long is it before she hears the splash?
30. 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?
31. 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?

Feb 23, 2012

Sep 15, 2014

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