# 10.9: Comparing Methods for Solving Quadratics

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

**Practice**Comparing Methods for Solving Quadratics

What if you and a friend started running from the same spot. You ran west and your friend ran south. After two hours, you had run 10 miles. The distance at that point between you and your friend was three times the distance your friend ran plus 2 miles? How could you determine how many miles your friend ran? After completing this Concept, you'll be able to solve real-world quadratic equation problems like this one.

### Watch This

CK-12 Foundation: 1009S Solving Real-World Problems Using Quadratic Equations

### Guidance

In mathematics, you’ll need to solve quadratic equations that describe application problems or that are part of more complicated problems. You’ve learned four ways of solving a quadratic equation:

- Factoring
- Taking the square root
- Quadratic formula
- Completing the square

Usually you’ll have to decide for yourself which method to use. However, here are some guidelines as to which methods are better in different situations.

**Factoring** is always best if the quadratic expression is easily factorable. It is always worthwhile to check if you can factor because this is the fastest method. Many expressions are not factorable so this method is not used very often in practice.

**Taking the square root** is best used when there is no \begin{align*}x-\end{align*}

**Quadratic formula** is the method that is used most often for solving a quadratic equation. When solving directly by taking square root and factoring does not work, this is the method that most people prefer to use.

**Completing the square** can be used to solve any quadratic equation. This is usually not any better than using the quadratic formula (in terms of difficult computations), but it is very useful if you need to rewrite a quadratic function in vertex form. It’s also used to rewrite the equations of circles, ellipses and hyperbolas in standard form (something you’ll do in algebra II, trigonometry, physics, calculus, and beyond).

If you are using factoring or the quadratic formula, make sure that the equation is in standard form.

#### Example A

*Solve each quadratic equation.*

a) \begin{align*}x^2 - 4x - 5 = 0\end{align*}

b) \begin{align*}x^2 = 8\end{align*}

c) \begin{align*}-4x^2 + x = 2\end{align*}

d) \begin{align*}25x^2 - 9 = 0\end{align*}

e) \begin{align*}3x^2 = 8x\end{align*}

**Solution**

a) This expression if easily factorable so we can factor and apply the zero-product property:

\begin{align*}& \text{Factor:} && (x - 5)(x + 1) = 0\\
& \text{Apply zero-product property:} && x - 5 = 0 \quad \text{and} \quad x + 1 = 0\\
& \text{Solve:} && x = 5 \qquad \ \ \text{and} \quad x = -1\end{align*}

**Answer:** \begin{align*}x = 5\end{align*}

b) Since the expression is missing the \begin{align*}x\end{align*}

Take the square root of both sides: \begin{align*}x = \sqrt{8}\end{align*}

**Answer:** \begin{align*}x = 2.83\end{align*}

c) Re-write the equation in standard form: \begin{align*}-4x^2 + x - 2 = 0\end{align*}

It is not apparent right away if the expression is factorable so we will use the quadratic formula:

\begin{align*}\text{Quadratic formula:} && x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\
\text{Plug in the values} \ a = -4, \ b = 1, \ c = -2: && x &= \frac{-1 \pm \sqrt{1^2 - 4(-4)(-2)}}{2(-4)}\\
\text{Simplify:} && x &= \frac{-1 \pm \sqrt{1 - 32}}{-8} = \frac{-1 \pm \sqrt{-31}}{-8}\end{align*}

**Answer:** no real solution

d) This problem can be solved easily either with factoring or taking the square root. Let’s take the square root in this case:

\begin{align*}\text{Add} \ 9 \ \text{to both sides of the equation:} && 25x^2 &= 9\\ \text{Divide both sides by} \ 25: && x^2 &= \frac{9}{25}\\ \text{Take the square root of both sides:} && x &= \sqrt{\frac{9}{25}} \ \text{and} \ x=-\sqrt{\frac{9}{25}}\\ \text{Simplify:} && x &= \frac{3}{5} \ \text{and} \ x=-\frac{3}{5}\end{align*}

**Answer:** \begin{align*}x = \frac{3}{5}\end{align*} and \begin{align*}x = -\frac{3}{5}\end{align*}

e) \begin{align*}\text{Re-write the equation in standard form:} && 3x^2 - 8x &= 0\\ \text{Factor out common} \ x \ \text{term:} && x(3x-8) &= 0\\ \text{Set both terms to zero:} && x &= 0 \ \text{and} \ 3x = 8\\ \text{Solve:} && x &= 0 \ \text{and} \ x = \frac{8}{3} = 2.67\end{align*}

**Answer:** \begin{align*}x = 0\end{align*} and \begin{align*}x = 2.67\end{align*}

**Solving Real-World Problems by Completing the Square**

In the last section you learned that an object that is dropped falls under the influence of gravity. The equation for its height with respect to time is given by \begin{align*}y = \frac{1}{2}gt^2 + y_0\end{align*}, where \begin{align*}y_0\end{align*} represents the initial height of the object and \begin{align*}g\end{align*} is the coefficient of gravity on earth, which equals \begin{align*}-9.8 \ m/s^2\end{align*} or \begin{align*}-32 \ ft/s^2\end{align*}.

On the other hand, if an object is thrown straight up or straight down in the air, it has an initial vertical velocity. This term is usually represented by the notation \begin{align*}v_{0y}\end{align*}. Its value is positive if the object is thrown up in the air and is negative if the object is thrown down. The equation for the height of the object in this case is

\begin{align*}y = \frac{1}{2}gt^2 + v_{0y}t + y_0\end{align*}

Plugging in the appropriate value for \begin{align*}g\end{align*} turns this equation into

\begin{align*}y = -4.9t^2 + v_{0y}t + y_0\end{align*} if you wish to have the height in meters

\begin{align*}y = -16t^2 + v_{0y}t + y_0\end{align*} if you wish to have the height in feet

#### Example B

*An arrow is shot straight up from a height of 2 meters with a velocity of 50 m/s.*

a) *How high will the arrow be 4 seconds after being shot? After 8 seconds?*

b) *At what time will the arrow hit the ground again?*

c) *What is the maximum height that the arrow will reach and at what time will that happen?*

**Solution**

Since we are given the velocity in m/s, use: \begin{align*}y = -4.9t^2 + v_{0y}t + y_0\end{align*}

We know \begin{align*}v_{0y} = 50 \ m/s\end{align*} and \begin{align*}y_0 = 2\end{align*} meters so: \begin{align*}y = -4.9t^2 + 50t + 2\end{align*}

a) To find how high the arrow will be 4 seconds after being shot we plug in \begin{align*}t = 4\end{align*}:

\begin{align*}y & = -4.9(4)^2 + 50(4) + 2\\ & = -4.9(16) + 200 + 2 = \underline{\underline{123.6 \ meters}}\end{align*}

we plug in \begin{align*}t = 8\end{align*}:

\begin{align*}y & = -4.9(8)^2 + 50(8) + 2\\ & = -4.9(64) + 400 + 2 = \underline{\underline{88.4 \ meters}}\end{align*}

b) The height of the arrow on the ground is \begin{align*}y = 0\end{align*}, so: \begin{align*}0 = -4.9t^2 + 50t + 2\end{align*}

Solve for \begin{align*}t\end{align*} by completing the square:

\begin{align*}-4.9t^2 + 50t & = -2\\ -4.9(t^2 - 10.2t) & = -2\\ t^2 - 10.2t & = 0.41\\ t^2 - 2(5.1)t + (5.1)^2 & = 0.41 + (5.1)^2\\ (t - 5.1)^2 & = 26.43\\ t - 5.1 & = 5.14 \ \text{and} \ t - 5.1 = -5.14\\ t & = \underline{\underline{10.2}} \ sec \ \text{and} \ t = -0.04 \ sec\end{align*}

**The arrow will hit the ground about 10.2 seconds after it is shot.**

c) If we graph the height of the arrow with respect to time we would get an upside down parabola \begin{align*}(a < 0)\end{align*}. The maximum height and the time when this occurs is really the vertex of this parabola: \begin{align*}(t, h)\end{align*}.

\begin{align*}& \text{We re-write the equation in vertex form:} && y = -4.9t^2 + 50t + 2\\ & && y - 2 = -4.9t^2 + 50t\\ & && y - 2 = -4.9(t^2 - 10.2t)\\ & \text{Complete the square:} && y - 2 - 4.9(5.1)^2 = -4.9 \left( t^2 - 10.2t + (5.1)^2 \right)\\ & && y - 129.45 = -4.9(t - 5.1)^2\end{align*}

The vertex is at (5.1, 129.45). In other words, **when \begin{align*}t = 5.1 \ seconds\end{align*}, the height is \begin{align*}y = 129 \ meters\end{align*}.**

Another type of application problem that can be solved using quadratic equations is one where two objects are moving away from each other in perpendicular directions. Here is an example of this type of problem.

#### Example C

*Two cars leave an intersection. One car travels north; the other travels east. When the car traveling north had gone 30 miles, the distance between the cars was 10 miles more than twice the distance traveled by the car heading east. Find the distance between the cars at that time.*

**Solution**

Let \begin{align*}x =\end{align*} the distance traveled by the car heading east

Then \begin{align*}2x + 10 =\end{align*} the distance between the two cars

Let’s make a sketch:

We can use the Pythagorean Theorem to find an equation for \begin{align*}x\end{align*}:

\begin{align*}x^2 + 30^2 = (2x + 10)^2\end{align*}

Expand parentheses and simplify:

\begin{align*}x^2 + 900 & = 4x^2 + 40x + 100\\ 800 & = 3x^2 + 40x\end{align*}

Solve by completing the square:

\begin{align*}\frac{800}{3} &= x^2 + \frac{40}{3}x\\ \frac{800}{3} + \left ( \frac{20}{3} \right )^2 &= x^2 + 2\left ( \frac{20}{3} \right )x + \left ( \frac{20}{3} \right )^2\\ \frac{2800}{9} &= \left ( x + \frac{20}{3} \right )^2\\ x + \frac{20}{3} &= 17.6 \ \text{and} \ x + \frac{20}{3} =-17.6\\ x &= 11 \ \text{and} \ x = -24.3\end{align*}

Since only positive distances make sense here, the distance between the two cars is: \begin{align*}2(11) + 10 = 32 \ miles\end{align*}

**Solve Applications Using Quadratic Functions by any Method**

Here is an application problem that arises from number relationships and geometry applications.

#### Example D

*The product of two positive consecutive integers is 156. Find the integers.*

**Solution**

**Define:** Let \begin{align*}x =\end{align*} the smaller integer

Then \begin{align*}x + 1 =\end{align*} the next integer

**Translate:** The product of the two numbers is 156. We can write the equation:

\begin{align*}x(x + 1) = 156\end{align*}

**Solve:**

\begin{align*}x^2 + x & = 156\\ x^2 + x - 156 & = 0\end{align*}

Apply the quadratic formula with: \begin{align*}a = 1, \ b = 1, \ c = -156\end{align*}

\begin{align*}x & = \frac{-1 \pm \sqrt{1^2 - 4(1)(-156)}}{2(1)}\\ x & = \frac{-1 \pm \sqrt{625}}{2} = \frac{-1 \pm 25}{2} \\ x & = \frac{-1 + 25}{2} \quad \text{and} \quad x = \frac{-1 - 25}{2}\\ x & = \frac{24}{2} = 12 \quad \text{and} \quad x = \frac{-26}{2} = -13\end{align*}

Since we are looking for positive integers, we want \begin{align*}x = 12\end{align*}. So the numbers are **12 and 13.**

**Check:** \begin{align*}12 \times 13 = 156\end{align*}. The answer checks out.

#### Example E

*Suzie wants to build a garden that has three separate rectangular sections. She wants to fence around the whole garden and between each section as shown. The plot is twice as long as it is wide and the total area is* \begin{align*}200 ft^2\end{align*}. *How much fencing does Suzie need?*

**Solution**

**Define:** Let \begin{align*}x =\end{align*} the width of the plot

Then \begin{align*}2x =\end{align*} the length of the plot

**Translate:** area of a rectangle is \begin{align*}A = \text{length} \times \text{width}\end{align*}, so

\begin{align*}x(2x) = 200\end{align*}

**Solve:** \begin{align*}2x^2 = 200\end{align*}

Solve by taking the square root:

\begin{align*}x^2 & = 100\\ x & = \sqrt{100} \ \text{and} \ x = - \sqrt{100}\\ x & = 10 \ \text{and} \ x = - 10\end{align*}

We take \begin{align*}x = 10\end{align*} since only positive dimensions make sense.

The plot of land is \begin{align*}10 \ feet \times 20 \ feet\end{align*}.

To fence the garden the way Suzie wants, we need 2 lengths and 4 widths \begin{align*}= 2(20) + 4(10) = 80\end{align*} **feet of fence.**

**Check:** \begin{align*}10 \times 20 = 200 \ ft^2\end{align*} and \begin{align*}2(20) + 4(10) = 80 \ feet\end{align*}. The answer checks out.

Watch this video for help with the Examples above.

CK-12 Foundation: 1009 Solving Real-World Problems Using Quadratic Equations

### Vocabulary

- For a
**quadratic equation**in standard form, \begin{align*}ax^2 + bx + c = 0\end{align*}, the**quadratic formula**looks like this:

\begin{align*}x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\end{align*}

- A
has the form \begin{align*}a^2+2(ab)+b^2\end{align*}, which factors into \begin{align*}(a+b)^2\end{align*}.*perfect square trinomial*

### Guided Practice

1. *An isosceles triangle is enclosed in a square so that its base coincides with one of the sides of the square and the tip of the triangle touches the opposite side of the square. If the area of the triangle is* \begin{align*}20 \ in^2\end{align*} *what is the length of one side of the square?*

2. *The length of a rectangular pool is 10 meters more than its width. The area of the pool is 875 square meters. Find the dimensions of the pool.*

**Solutions:**

1. **Draw a sketch:**

**Define:** Let \begin{align*}x =\end{align*} base of the triangle

Then \begin{align*}x =\end{align*} height of the triangle

**Translate:** Area of a triangle is \begin{align*} \frac{1}{2} \times base \times height\end{align*}, so \begin{align*} \frac{1}{2} \cdot x \cdot x = 20\end{align*}

**Solve:** \begin{align*}\frac{1}{2}x^2 = 20\end{align*}

Solve by taking the square root:

\begin{align*}x^2 & = 40\\ x & = \sqrt{40} \ \text{and} \ x = - \sqrt{40}\\ x & = 6.32 \ \text{and} \ x = -6.32\end{align*}

The side of the square is **6.32 inches.** That means the area of the square is \begin{align*}(6.32)^2 = 40 \ in^2\end{align*}, twice as big as the area of the triangle.

**Check:** It makes sense that the area of the square will be twice that of the triangle. If you look at the figure you can see that you could fit two triangles inside the square.

2. **Draw a sketch:**

**Define:** Let \begin{align*}x =\end{align*} the width of the pool

Then \begin{align*}x + 10 =\end{align*} the length of the pool

**Translate:** The area of a rectangle is \begin{align*}A = \text{length} \times \text{width}\end{align*}, so we have \begin{align*}x(x +10) = 875\end{align*}.

**Solve:**

\begin{align*}x^2 + 10x &= 875\\ x^2 + 10x - 875 &= 0\end{align*}

Apply the quadratic formula with \begin{align*}a = 1, \ b = 10\end{align*} and \begin{align*}c = -875\end{align*}

\begin{align*}x & = \frac{-10 \pm \sqrt{(10)^2 - 4(1)(-875)}}{2(1)}\\ x & = \frac{-10 \pm \sqrt{100 + 3500}}{2}\\ x & = \frac{-10 \pm \sqrt{3600}}{2} = \frac{-10 \pm 60}{2}\\ x & = \frac{-10 + 60}{2} \ \text{and} \ x = \frac{-10 - 60}{2}\\ x & = \frac{50}{2} = 25 \ \text{and} \ x = \frac{-70}{2} = -35\end{align*}

Since the dimensions of the pool should be positive, we want \begin{align*}x = 25 \ meters\end{align*}. So the pool is \begin{align*}25 \ meters \times 35 \ meters\end{align*}.

**Check:** \begin{align*}25 \times 35 = 875 \ m^2\end{align*}. The answer checks out.

### Practice

Solve the following quadratic equations using the method of your choice.

- \begin{align*}x^2 - x = 6\end{align*}
- \begin{align*}x^2 - 12 = 0\end{align*}
- \begin{align*}-2x^2 + 5x - 3 = 0\end{align*}
- \begin{align*}x^2 + 7x - 18 = 0\end{align*}
- \begin{align*}3x^2 + 6x = - 10\end{align*}
- \begin{align*}-4x^2 + 4000x = 0\end{align*}
- \begin{align*}-3x^2 + 12x + 1 = 0\end{align*}
- \begin{align*}x^2 + 6x + 9 = 0\end{align*}
- \begin{align*}81x^2 + 1 = 0\end{align*}
- \begin{align*}-4x^2 + 4x = 9\end{align*}
- \begin{align*}36x^2 - 21 = 0\end{align*}
- \begin{align*}x^2 - 2x - 3 = 0\end{align*}
- The product of two consecutive integers is 72. Find the two numbers.
- The product of two consecutive odd integers is 1 less than 3 times their sum. Find the integers.
- The length of a rectangle exceeds its width by 3 inches. The area of the rectangle is 70 square inches, find its dimensions.
- Angel wants to cut off a square piece from the corner of a rectangular piece of plywood. The larger piece of wood is \begin{align*}4 \ feet \times 8 \ feet\end{align*} and the cut off part is \begin{align*}\frac{1}{3}\end{align*} of the total area of the plywood sheet. What is the length of the side of the square?
- Mike wants to fence three sides of a rectangular patio that is adjacent the back of his house. The area of the patio is \begin{align*}192 \ ft^2\end{align*} and the length is 4 feet longer than the width. Find how much fencing Mike will need.
- Sam throws an egg straight down from a height of 25 feet. The initial velocity of the egg is 16 ft/sec. How long does it take the egg to reach the ground?
- Amanda and Dolvin leave their house at the same time. Amanda walks south and Dolvin bikes east. Half an hour later they are 5.5 miles away from each other and Dolvin has covered three miles more than the distance that Amanda covered. How far did Amanda walk and how far did Dolvin bike?

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### Image Attributions

Here you'll learn how to choose the best method (factoring, taking the square root, using the quadratic formula, or completing the square) to solve real-world applications involving quadratic equations.