## Primary Study Cards

Terms and Conditions

## 9.5 Solve Applications of Quadratic Equations

Learning objectives.

By the end of this section, you will be able to:

• Solve applications modeled by quadratic equations

## Be Prepared 9.13

Before you get started, take this readiness quiz.

The sum of two consecutive odd numbers is −100. Find the numbers. If you missed this problem, review Example 2.18 .

## Be Prepared 9.14

Solve: 2 x + 1 + 1 x − 1 = 1 x 2 − 1 . 2 x + 1 + 1 x − 1 = 1 x 2 − 1 . If you missed this problem, review Example 7.35 .

## Be Prepared 9.15

Find the length of the hypotenuse of a right triangle with legs 5 inches and 12 inches. If you missed this problem, review Example 2.34 .

Solve Applications Modeled by Quadratic Equations

We solved some applications that are modeled by quadratic equations earlier, when the only method we had to solve them was factoring. Now that we have more methods to solve quadratic equations, we will take another look at applications.

Let’s first summarize the methods we now have to solve quadratic equations.

## Methods to Solve Quadratic Equations

• Square Root Property
• Completing the Square

As you solve each equation, choose the method that is most convenient for you to work the problem. As a reminder, we will copy our usual Problem-Solving Strategy here so we can follow the steps.

## Use a Problem-Solving Strategy.

• Step 1. Read the problem. Make sure all the words and ideas are understood.
• Step 2. Identify what we are looking for.
• Step 3. Name what we are looking for. Choose a variable to represent that quantity.
• Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
• Step 5. Solve the equation using algebra techniques.
• Step 6. Check the answer in the problem and make sure it makes sense.
• Step 7. Answer the question with a complete sentence

We have solved number applications that involved consecutive even and odd integers, by modeling the situation with linear equations. Remember, we noticed each even integer is 2 more than the number preceding it. If we call the first one n , then the next one is n + 2. The next one would be n + 2 + 2 or n + 4. This is also true when we use odd integers. One set of even integers and one set of odd integers are shown below.

Some applications of odd or even consecutive integers are modeled by quadratic equations. The notation above will be helpful as you name the variables.

## Example 9.35

The product of two consecutive odd integers is 195. Find the integers.

## Try It 9.69

The product of two consecutive odd integers is 99. Find the integers.

## Try It 9.70

The product of two consecutive even integers is 168. Find the integers.

We will use the formula for the area of a triangle to solve the next example.

## Area of a Triangle

For a triangle with base, b , and height, h , the area, A , is given by the formula A = 1 2 b h . A = 1 2 b h .

Recall that when we solve geometric applications, it is helpful to draw the figure.

## Example 9.36

An architect is designing the entryway of a restaurant. She wants to put a triangular window above the doorway. Due to energy restrictions, the window can only have an area of 120 square feet and the architect wants the base to be 4 feet more than twice the height. Find the base and height of the window.

## Try It 9.71

Find the base and height of a triangle whose base is four inches more than six times its height and has an area of 456 square inches.

## Try It 9.72

If a triangle that has an area of 110 square feet has a base that is two feet less than twice the height, what is the length of its base and height?

In the two preceding examples, the number in the radical in the Quadratic Formula was a perfect square and so the solutions were rational numbers. If we get an irrational number as a solution to an application problem, we will use a calculator to get an approximate value.

We will use the formula for the area of a rectangle to solve the next example.

## Area of a Rectangle

For a rectangle with length, L , and width, W , the area, A , is given by the formula A = LW .

## Example 9.37

Mike wants to put 150 square feet of artificial turf in his front yard. This is the maximum area of artificial turf allowed by his homeowners association. He wants to have a rectangular area of turf with length one foot less than 3 times the width. Find the length and width. Round to the nearest tenth of a foot.

## Try It 9.73

The length of a 200 square foot rectangular vegetable garden is four feet less than twice the width. Find the length and width of the garden, to the nearest tenth of a foot.

## Try It 9.74

A rectangular tablecloth has an area of 80 square feet. The width is 5 feet shorter than the length.What are the length and width of the tablecloth to the nearest tenth of a foot.?

The Pythagorean Theorem gives the relation between the legs and hypotenuse of a right triangle. We will use the Pythagorean Theorem to solve the next example.

## Pythagorean Theorem

In any right triangle, where a and b are the lengths of the legs, and c is the length of the hypotenuse, a 2 + b 2 = c 2 .

## Example 9.38

Rene is setting up a holiday light display. He wants to make a ‘tree’ in the shape of two right triangles, as shown below, and has two 10-foot strings of lights to use for the sides. He will attach the lights to the top of a pole and to two stakes on the ground. He wants the height of the pole to be the same as the distance from the base of the pole to each stake. How tall should the pole be?

## Try It 9.75

The sun casts a shadow from a flag pole. The height of the flag pole is three times the length of its shadow. The distance between the end of the shadow and the top of the flag pole is 20 feet. Find the length of the shadow and the length of the flag pole. Round to the nearest tenth.

## Try It 9.76

The distance between opposite corners of a rectangular field is four more than the width of the field. The length of the field is twice its width. Find the distance between the opposite corners. Round to the nearest tenth.

The height of a projectile shot upward from the ground is modeled by a quadratic equation. The initial velocity, v 0 , propels the object up until gravity causes the object to fall back down.

## Projectile motion

The height in feet, h , of an object shot upwards into the air with initial velocity, v 0 v 0 , after t t seconds is given by the formula

We can use this formula to find how many seconds it will take for a firework to reach a specific height.

## Example 9.39

A firework is shot upwards with initial velocity 130 feet per second. How many seconds will it take to reach a height of 260 feet? Round to the nearest tenth of a second.

## Try It 9.77

An arrow is shot from the ground into the air at an initial speed of 108 ft/s. Use the formula h = −16 t 2 + v 0 t to determine when the arrow will be 180 feet from the ground. Round the nearest tenth.

## Try It 9.78

A man throws a ball into the air with a velocity of 96 ft/s. Use the formula h = −16 t 2 + v 0 t to determine when the height of the ball will be 48 feet. Round to the nearest tenth.

We have solved uniform motion problems using the formula D = rt in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

The formula D = rt assumes we know r and t and use them to find D . If we know D and r and need to find t , we would solve the equation for t and get the formula t = D r . t = D r .

Some uniform motion problems are also modeled by quadratic equations.

## Example 9.40

Professor Smith just returned from a conference that was 2,000 miles east of his home. His total time in the airplane for the round trip was 9 hours. If the plane was flying at a rate of 450 miles per hour, what was the speed of the jet stream?

This is a uniform motion situation. A diagram will help us visualize the situation.

## Try It 9.79

MaryAnne just returned from a visit with her grandchildren back east . The trip was 2400 miles from her home and her total time in the airplane for the round trip was 10 hours. If the plane was flying at a rate of 500 miles per hour, what was the speed of the jet stream?

## Try It 9.80

Gerry just returned from a cross country trip. The trip was 3000 miles from his home and his total time in the airplane for the round trip was 11 hours. If the plane was flying at a rate of 550 miles per hour, what was the speed of the jet stream?

Work applications can also be modeled by quadratic equations. We will set them up using the same methods we used when we solved them with rational equations.We’ll use a similar scenario now.

## Example 9.41

The weekly gossip magazine has a big story about the presidential election and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 12 hours more than Press #2 to do the job and when both presses are running they can print the job in 8 hours. How long does it take for each press to print the job alone?

This is a work problem. A chart will help us organize the information.

We are looking for how many hours it would take each press separately to complete the job.

## Try It 9.81

The weekly news magazine has a big story naming the Person of the Year and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 6 hours more than Press #2 to do the job and when both presses are running they can print the job in 4 hours. How long does it take for each press to print the job alone?

## Try It 9.82

Erlinda is having a party and wants to fill her hot tub. If she only uses the red hose it takes 3 hours more than if she only uses the green hose. If she uses both hoses together, the hot tub fills in 2 hours. How long does it take for each hose to fill the hot tub?

Access these online resources for additional instruction and practice with solving applications modeled by quadratic equations.

## Section 9.5 Exercises

Practice makes pefect.

In the following exercises, solve using any method.

The product of two consecutive odd numbers is 255. Find the numbers.

The product of two consecutive even numbers is 360. Find the numbers.

The product of two consecutive even numbers is 624. Find the numbers.

The product of two consecutive odd numbers is 1,023. Find the numbers.

The product of two consecutive odd numbers is 483. Find the numbers.

The product of two consecutive even numbers is 528. Find the numbers.

In the following exercises, solve using any method. Round your answers to the nearest tenth, if needed.

A triangle with area 45 square inches has a height that is two less than four times the base Find the base and height of the triangle.

The base of a triangle is six more than twice the height. The area of the triangle is 88 square yards. Find the base and height of the triangle.

The area of a triangular flower bed in the park has an area of 120 square feet. The base is 4 feet longer that twice the height. What are the base and height of the triangle?

A triangular banner for the basketball championship hangs in the gym. It has an area of 75 square feet. What is the length of the base and height , if the base is two-thirds of the height?

The length of a rectangular driveway is five feet more than three times the width. The area is 50 square feet. Find the length and width of the driveway.

A rectangular lawn has area 140 square yards. Its width that is six less than twice the length. What are the length and width of the lawn?

A rectangular table for the dining room has a surface area of 24 square feet. The length is two more feet than twice the width of the table. Find the length and width of the table.

The new computer has a surface area of 168 square inches. If the the width is 5.5 inches less that the length, what are the dimensions of the computer?

The hypotenuse of a right triangle is twice the length of one of its legs. The length of the other leg is three feet. Find the lengths of the three sides of the triangle.

The hypotenuse of a right triangle is 10 cm long. One of the triangle’s legs is three times the length of the other leg. Find the lengths of the two legs of the triangle. Round to the nearest tenth.

A rectangular garden will be divided into two plots by fencing it on the diagonal. The diagonal distance from one corner of the garden to the opposite corner is five yards longer than the width of the garden. The length of the garden is three times the width. Find the length of the diagonal of the garden.

Nautical flags are used to represent letters of the alphabet. The flag for the letter, O consists of a yellow right triangle and a red right triangle which are sewn together along their hypotenuse to form a square. The hypotenuse of the two triangles is three inches longer than a side of the flag. Find the length of the side of the flag.

Gerry plans to place a 25-foot ladder against the side of his house to clean his gutters. The bottom of the ladder will be 5 feet from the house.How far up the side of the house will the ladder reach?

John has a 10-foot piece of rope that he wants to use to support his 8-foot tree. How far from the base of the tree should he secure the rope?

A firework rocket is shot upward at a rate of 640 ft/sec. Use the projectile formula h = −16 t 2 + v 0 t to determine when the height of the firework rocket will be 1200 feet.

An arrow is shot vertically upward at a rate of 220 feet per second. Use the projectile formula h = −16 t 2 + v 0 t , to determine when height of the arrow will be 400 feet.

A bullet is fired straight up from a BB gun with initial velocity 1120 feet per second at an initial height of 8 feet. Use the formula h = −16 t 2 + v 0 t + 8 to determine how many seconds it will take for the bullet to hit the ground. (That is, when will h = 0?)

A stone is dropped from a 196-foot platform. Use the formula h = −16 t 2 + v 0 t + 196 to determine how many seconds it will take for the stone to hit the ground. (Since the stone is dropped, v 0 = 0.)

The businessman took a small airplane for a quick flight up the coast for a lunch meeting and then returned home. The plane flew a total of 4 hours and each way the trip was 200 miles. What was the speed of the wind that affected the plane which was flying at a speed of 120 mph?

The couple took a small airplane for a quick flight up to the wine country for a romantic dinner and then returned home. The plane flew a total of 5 hours and each way the trip was 300 miles. If the plane was flying at 125 mph, what was the speed of the wind that affected the plane?

Roy kayaked up the river and then back in a total time of 6 hours. The trip was 4 miles each way and the current was difficult. If Roy kayaked at a speed of 5 mph, what was the speed of the current?

Rick paddled up the river, spent the night camping, and then paddled back. He spent 10 hours paddling and the campground was 24 miles away. If Rick kayaked at a speed of 5 mph, what was the speed of the current?

Two painters can paint a room in 2 hours if they work together. The less experienced painter takes 3 hours more than the more experienced painter to finish the job. How long does it take for each painter to paint the room individually?

Two gardeners can do the weekly yard maintenance in 8 minutes if they work together. The older gardener takes 12 minutes more than the younger gardener to finish the job by himself. How long does it take for each gardener to do the weekly yard maintainence individually?

It takes two hours for two machines to manufacture 10,000 parts. If Machine #1 can do the job alone in one hour less than Machine #2 can do the job, how long does it take for each machine to manufacture 10,000 parts alone?

Sully is having a party and wants to fill his swimming pool. If he only uses his hose it takes 2 hours more than if he only uses his neighbor’s hose. If he uses both hoses together, the pool fills in 4 hours. How long does it take for each hose to fill the pool?

## Writing Exercises

Make up a problem involving the product of two consecutive odd integers.

ⓐ Start by choosing two consecutive odd integers. What are your integers?

ⓑ What is the product of your integers?

ⓒ Solve the equation n ( n + 2) = p , where p is the product you found in part (b).

ⓓ Did you get the numbers you started with?

Make up a problem involving the product of two consecutive even integers.

ⓐ Start by choosing two consecutive even integers. What are your integers?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

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• Authors: Lynn Marecek, Andrea Honeycutt Mathis
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• Book title: Intermediate Algebra 2e
• Publication date: May 6, 2020
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## Concept Quizzes

• Distributive Property - Multiple Terms
• Quadratic Equations - Word Problems
• Quadratic Equations - Problem Solving

## Challenge Quizzes

• Quadratic Equations: Level 2 Challenges
• Quadratic Equations: Level 3 Challenges
• Quadratic Equations: Level 4 Challenges
• Multiple Terms

What it is, what it does, and how to use it

## What is a quadratic equation?

A quadratic equation is an equation that can be written as ax ² + bx + c where a ≠ 0 . In other words, a quadratic equation must have a squared term as its highest power.

$$y = 5x^2 + 2x + 5 \\ y = 11x^2 + 22 \\ y = x^2 - 4x +5 \\ y = -x^2 + + 5$$

## Non Examples

$$y = 11x + 22 \\ y = x^3 -x^2 +5x +5 \\ y = 2x^3 -4x^2 \\ y = -x^4 + 5$$

## Ok, but what is a 'solution'?

Well a solution can be thought in two ways:

## Example of the quadratic formula to solve an equation

Use the formula to solve theQuadratic Equation: $$y = x^2 + 2x + 1$$ .

Just substitute a,b, and c into the general formula:

$$a = 1 \\ b = 2 \\ c = 1$$

Below is a picture representing the graph of y = x² + 2x + 1 and its solution.

A catchy way to remember the quadratic formula is this song (pop goes the weasel).

## Practice Problems

Use the quadratic formula to find the solutions to the following equation: y = x² − 2x + 1 and its solution.

• b = − 2

Use the quadratic formula to find the solutions to the following equation: y = x² − x − 2 and its solution .

In this quadratic equation, y = x² − x − 2 and its solution:

• b = − 1
• c = − 2

Use the quadratic formula to find the solutions to the following equation: y = x² − 1 and its solution.

In this quadratic equation, y = x² − 1 and its solution:

• c = −1

Calculate the solutions of the the quadratic equation below by using the quadratic formula : y = x² + 2x − 3 and its solution.

In this quadratic equation, y = x² + 2x − 3 and its solution:

• c = −3

Below is a picture of the graph of the quadratic equation and its two solutions.

Use the quadratic formula to find the solutions to the following equation: y = x² + 4x − 5 and its solution.

In this quadratic equation, y = x² + 4x − 5 and its solution:

• c = −5

Use the quadratic formula to find the solutions to the following equation: y = x² − 4x + 5 and its solution.

In this quadratic equation, y = x² − 4x + 5 and its solution:

• b = −4

Below is a picture of this quadratic's graph.

• Discriminant
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Quadratic Formula Discriminant Disc. & Graphs

## When would I use the Quadratic Formula?

You can use the Quadratic Formula any time you're trying to solve a quadratic equation — as long as that equation is in the form "(a quadratic expression) that is set equal to zero".

Often, the simplest way to solve " ax 2 + bx + c = 0 " for the value of x is to factor the quadratic , set each factor equal to zero, and then solve each factor. But sometimes the quadratic is too messy, or it doesn't factor at all, or, heck, maybe you just don't feel like factoring. While factoring is not always going to be successful, the Quadratic Formula can always find the answers for you.

Content Continues Below

## MathHelp.com

The Quadratic Formula uses the " a ", " b ", and " c " from " ax 2 + bx + c ", where " a ", " b ", and " c " are just numbers; they are the "numerical coefficients" of the quadratic equation they've given you to solve.

The Quadratic Formula is derived from the process of completing the square, and is formally stated as:

## What is the Quadratic Formula?

The Quadratic Formula is a rule that says that, in any equation of the form ax 2 + bx + c = 0 , the solution x -values of the equation are given by:

## How do I use the Quadratic Formula?

To use the Quadratic Formula, you must:

Arrange the terms in the (equation) in decreasing order (so squared term first, then the x -term, and finally the linear term).

Pull out the numerical parts of each of these terms, which are the " a ", " b ", and " c " of the Formula.

Plug these numbers into the formula.

Advisories: The " 2 a " in the denominator of the Formula is underneath everything above, not just the square root. And it's a " 2 a " under there, not just a plain " 2 ". Make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them back in" on your test, and you'll mess yourself up. Remember that " b 2 " means "the square of ALL of b , including its sign", so don't leave b 2 being negative, even if b is negative, because the square of a negative is a positive.

In other words, don't be sloppy and don't try to take shortcuts, because it will only hurt you in the long run. Trust me on this!

## What is an example of using the Quadratic Formula?

Solve x 2 + 3 x − 4 = 0.

This quadratic happens to factor, which I can use to confirm what I get from the Quadratic Formula. The Formula should give me the same answers.

x 2 + 3 x − 4 = ( x + 4)( x − 1) = 0

...so I already know that the solutions are x = −4 and x = 1 .

Now, what would my solution look like in the Quadratic Formula? Using a = 1 , b = 3 , and c = −4 , my solution process looks like this:

So, as expected, the solution is x = −4 , x = 1 .

For this particular quadratic equation, factoring would probably be the faster method. But the Quadratic Formula is a plug-n-chug method that will always work. Having "brain freeze" on a test and can't factor worth a darn? Use the plug-n-chug Formula; it'll always take care of you!

## How does the Quadratic Formula relate to x -intercepts?

The solutions to the quadratic equation, as provided by the Quadratic Formula, are the x -intercepts of the corresponding graphed parabola.

How? Well, when y = 0 , you're on the x -axis. The x -intercepts of the graph are where the parabola crosses the x -axis. You're applying the Quadratic Formula to the equation ax 2 + bx + c = y , where y is set equal to zero.

Looking at the above example, there were two solutions for the equation x 2 + 3 x − 4 = 0 . This tells us that there must then be two x -intercepts on the graph. Graphing, we get the curve below:

As you can see, the x -intercepts (the red dots above) match the solutions, crossing the x -axis at x = −4 and x = 1 . This shows the connection between graphing and solving: When you are solving "(quadratic) = 0 ", you are finding the x -intercepts of the graph. This can be useful if you have a graphing calculator, because you can use the Quadratic Formula (when necessary) to solve a quadratic, and then use your graphing calculator to make sure that the displayed x -intercepts have the same decimal values as do the solutions provided by the Quadratic Formula.

Note, however, that the calculator's display of the graph will probably have some pixel-related round-off error, so you'd be checking to see that the computed and graphed values were reasonably close; don't expect an exact match.

## Solve 2 x 2 − 4 x − 3 = 0 . Round your answer to two decimal places, if necessary.

There are no factors of (2)(−3) = −6 that add up to −4 , so I know that this quadratic cannot be factored . I will apply the Quadratic Formula. In this case, a = 2, b = −4 , and c = −3 :

Then the answer is x = −0.58, x = 2.58 , rounded to two decimal places.

In general, no, you really shouldn't; the "solution" or "roots" or "zeroes" of a quadratic are usually required to be in the "exact" form of the answer. You can use the rounded form when graphing (if necessary), but "the answer(s)" from the Quadratic Formula should be written out in the (often messy) "exact" form.

In the example above, the exact form is the one with the square roots of ten in it. If you're wanting to graph the x -intercepts or needing to simplify the final answer in a word problem to be of a practical ("real world") form, then you can use the calculator's approximation. But unless you have a good reason to think that the answer is supposed to be a rounded answer, always go with the exact form.

Reinforcing the concept: Compare the solutions we found above for the equation 2 x 2 − 4 x − 3 = 0 with the x -intercepts of the graph:

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## Standard Form Of A Quadratic Equation

We have observed that a quadratic equation is an equation of the form

$$ax^2 + bx + c = 0, a \not = 0$$.

• $$a$$ is the coefficient of the quadratic term,
• $$b$$ is the coefficient of the linear term,
• $$c$$ is the constant term.

## Standard Form

The equation $$ax^2 + bc + c = 0$$ is the standard form of a quadratic equation.

## Sample Set A

Determine the values of $$a, b$$, and $$c$$.

## Example $$\PageIndex{1}$$

In the equation $$3x^2 + 5x + 2 = 0$$,

## Example $$\PageIndex{2}$$

In the equation $$12x^2-2x-1 = 0$$,

## Example $$\PageIndex{3}$$

In the equation $$2y^2 + 3 = 0$$,

$$b = 0$$ -> Because the equation could be wrirten $$2y^2 + 0y + 3 = 0$$

## Example $$\PageIndex{4}$$

In the equation $$-8y^2 + 11y = 0$$,

$$c = 0$$ -> Since $$-8y^2 + 11y + 0 = 0$$

## Example $$\PageIndex{5}$$

In the equation $$z^2 = z + 8$$,

When we write the equation in standard form, we get $$z^2 - z - 8 = 0$$

## Practice Set A

Determine the values of $$a, b$$, and $$c$$ in the following quadratic equations.

## Practice Problem $$\PageIndex{1}$$

$$4x^2 - 3x + 5 = 0$$

$$\begin{array}{flushleft} a &= 4\\ b &= -3\\ c &= 5 \end{array}$$

## Practice Problem $$\PageIndex{2}$$

$$3y^2 - 2y + 9 = 0$$

$$\begin{array}{flushleft} a &= 3\\ b &= -2\\ c &= 9 \end{array}$$

## Practice Problem $$\PageIndex{3}$$

$$x^2 - 5x - 1 = 0$$

$$\begin{array}{flushleft} a &= 1\\ b &= -5\\ c &= -1 \end{array}$$

## Practice Problem $$\PageIndex{4}$$

$$x^2 - 4 = 0$$

$$\begin{array}{flushleft} a &= 1\\ b &= 0\\ c &= -4 \end{array}$$

## Practice Problem $$\PageIndex{5}$$

$$x^2 - 2x = 0$$

$$\begin{array}{flushleft} a &= 1\\ b &= -2\\ c &= 0 \end{array}$$

## Practice Problem $$\PageIndex{6}$$

$$y^2 = 5y - 6$$

$$\begin{array}{flushleft} a &= 1\\ b &= -5\\ c &= 6 \end{array}$$

## Practice Problem $$\PageIndex{7}$$

$$2x^2 - 4x = -1$$

$$\begin{array}{flushleft} a &= 2\\ b &= -4\\ c &= 1 \end{array}$$

## Practice Problem $$\PageIndex{8}$$

$$5x - 3 = -3x^2$$

$$\begin{array}{flushleft} a &= 3\\ b &= 5\\ c &= -3 \end{array}$$

## Practice Problem $$\PageIndex{9}$$

$$2x - 11 - 3x^2 = 0$$

$$\begin{array}{flushleft} a &= -3\\ b &= 2\\ c &= -11 \end{array}$$

## Practice Problem $$\PageIndex{10}$$

$$y^2 = 0$$

$$\begin{array}{flushleft} a &= 1\\ b &= 0\\ c &= 0 \end{array}$$

The solutions to all quadratic equations depend only and completely on the values $$a, b$$ and $$c$$

When a quadratic equation is written in standard form so that the values $$a, b$$, and $$c$$ are readily determined, the equation can be solved using the quadratic formula . The values that satisfy the equation are found by substituting the values $$a, b$$, and $$c$$ into the formula

$$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Keep in mind that the plus or minus symbol, \pm, is just a shorthand way or denoting the two possibilities:

$$x = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a}$$ and $$x = \dfrac{-b - \sqrt{b^2 - 4ac}}{2a}$$

The quadratic formula can be derived by using the method of completing the square.

## Derivation Of The Quadratic Formula

Solve $$ax^2 + bx = -c = 0$$ for $$x$$ by completing the square.

Subtract $$c$$ from both sides.

$$ax^2 + bx = -c$$.

Divide both sides by $$a$$, the coefficient of $$x^2$$.

$$x^2 + \dfrac{b}{a}x = \dfrac{-c}{a}$$

Now we have the proper form to complete the square. Take one half the coefficient of $$x$$, square it, and add the result to both sides of the equation found in step 2.

a) $$\dfrac{1}{2} \cdot \dfrac{b}{a} = \dfrac{b}{2a}$$ is one half the coefficient of $$x$$.

b) $$(\dfrac{b}{2a})^2$$ is the square of one half the coefficient of $$x$$

$$x^2 + \dfrac{b}{a} + (\dfrac{b}{2a})^2 = \dfrac{-c}{a} + (\dfrac{b}{2a})^2$$

The left side of the equation is now a perfect square trinomial and can be factored. This gives us:

$$(x + \dfrac{b}{2a})^2 = \dfrac{-c}{a} + \dfrac{b^2}{4a^2}$$

Add the two fractions on the right side of the equation. The LCD $$= 4a^2$$.

$$\begin{array}{flushleft} (x + \dfrac{b}{2a})^2 &= \dfrac{-4ac}{4a^2} + \dfrac{b^2}{4a^2}\\ (x + \dfrac{b}{2a})^2 &= \dfrac{-4ac + b^2}{4a^2}\\ (x + \dfrac{b}{2a})^2 &= \dfrac{b^2 - 4ac}{4a^2} \end{array}$$

Solve for $$x$$ using the method of extraction of roots.

$$\begin{array}{flushleft} x + \dfrac{b}{2a} &= \pm \sqrt{\dfrac{b^2 - 4ac}{4a^2}}\\ x + \dfrac{b}{2a} &= \pm \dfrac{\sqrt{b^2 - 4ac}}{\sqrt{4a^2}} & \sqrt{4a^2} = |2a| = 2|a| = \pm 2a\\ x + \dfrac{b}{2a} &= \pm \dfrac{\sqrt{b^2 - 4ac}}{2a}\\ x &= -\dfrac{b}{2a} \pm \dfrac{\sqrt{b^2 - 4ac}}{2a} & \text{ Add these two fractions }\\ x &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{array}$$

## Example $$\PageIndex{6}$$

$$3x^2 + 5x + 2 = 0$$

1. Identify $$a, b$$, and $$c$$.

$$a = 3, b = 5, c = 2$$

3. Substitute.

$$\begin{array}{flushleft} x &= \dfrac{-5 \pm \sqrt{(5)^2 - 4(3)(2)}}{2(3)}\\ &= \dfrac{-5 \pm \sqrt{25 - 24}}{6}\\ &= \dfrac{-5 \pm \sqrt{1}}{6}\\ &= \dfrac{-5 + 1}{6} & -5 + 1 = -4 \text{ and } -5 - 1 = -6 &= \dfrac{-4}{6}, \dfrac{-6}{6}\\ x & \dfrac{-2}{3}, -1 \end{array}$$

Note: Since these roots are rational numbers, this equation could have been solved by factoring.

## Example $$\PageIndex{7}$$

$$12x^2 - 2x - 1 = 0$$

$$a = 12, b = -2, c = -1$$

$$\begin{array}{flushleft} x &= \dfrac{(-2) \pm \sqrt{(-2)^2 - 4(12)(-1)}}{2(12)}\\ &= \dfrac{2 \pm \sqrt{4 + 48}}{24} & \text{ Simplify }\\ &= \dfrac{2 \pm \sqrt{52}}{24} & \text{ Simplify }\\ &= \dfrac{2 \pm \sqrt{4 \cdot 13}}{24} & \text{ Simplify }\\ &= \dfrac{2 \pm 2\sqrt{13}}{24} & \text{ Reduce. Factor } 2 \text{ from the terms of the numerator.}\\ &= \dfrac{2(1 \pm \sqrt{13})}{24}\\ x &= \dfrac{1 \pm \sqrt{13}}{12} \end{array}$$

## Example $$\PageIndex{8}$$

$$2y^2 + 3 = 0$$

$$a = 2, b = 0, c = 3$$

$$\begin{array}{flushleft} x &= \dfrac{-0 \pm \sqrt{0^2 - 4(2)(3)}}{2(2)}\\ x &= \dfrac{0 \pm \sqrt{-24}}{4} \end{array}$$

This equation has no real number solution since we have obtained a negative number under the radical sign.

## Example $$\PageIndex{9}$$

$$-8x^2 + 11x = 0$$

$$a = -8, b = 11, c = 0$$

$$\begin{array}{flushleft} x &= \dfrac{-11 \pm \sqrt{11^2 - 4(-8)(0)}}{2(-8)}\\ &= \dfrac{-11 \pm \sqrt{121 - 0}}{-16} & \text{ Simplify }\\ &= \dfrac{-11 \pm \sqrt{121}}{-16} & \text{ Simplify }\\ &= \dfrac{-11 \pm 11}{-16} x &= 0, \dfrac{11}{8} \end{array}$$

## Example $$\PageIndex{10}$$

$$(3x + 1)(x - 4) = x^2 + x - 2$$

1. Write the equation in standard form.

$$\begin{array}{flushleft} 3x^2 - 11x - 4 &= x^2 + x - 2\\ 2x^2 - 12x - 2 &= 0\\ x^2 - 6x - 1 &= 0 \end{array}$$

2. Identify $$a, b$$, and $$c$$.

$$a = 1, b = -6, c = -1$$

4. Substitute.

$$\begin{array}{flushleft} x &= \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-1)}}{2(1)}\\ &= \dfrac{6 \pm \sqrt{36 + 4}}{2}\\ &= \dfrac{6 \pm \sqrt{40}}{2}\\ &= \dfrac{6 \pm \sqrt{4 \cdot 10}}{2}\\ &= \dfrac{6 \pm 2\sqrt{10}}{2}\\ &= \dfrac{2(3 \pm \sqrt{10}}{2} \end{array}$$

$$x = 3 \pm \sqrt{10}$$

## Practice Set B

Practice problem $$\pageindex{11}$$.

$$2x^2 + 3x - 7 = 0$$

$$x = \dfrac{-3 \pm \sqrt{65}}{4}$$

## Practice Problem $$\PageIndex{12}$$

$$5a^2 - 2a - 1 = 0$$

$$a = \dfrac{1 \pm \sqrt{6}}{5}$$

## Practice Problem $$\PageIndex{13}$$

$$6y^2 + 5 = 0$$

no real number solution

## Practice Problem $$\PageIndex{14}$$

$$-3m^2 + 2m = 0$$

Add texts here. Do not delete this text first.

For the following problems, solve the equations using the quadratic formula.

## Exercise $$\PageIndex{1}$$

$$x^2 - 2x - 3 = 0$$

$$x=3, −1$$

## Exercise $$\PageIndex{2}$$

$$x^2 + 5x + 6 = 0$$

## Exercise $$\PageIndex{3}$$

$$y^2 - 5y + 4 = 0$$

## Exercise $$\PageIndex{4}$$

$$a^2 + 4a - 21 = 0$$

## Exercise $$\PageIndex{5}$$

$$a^2 + 12a + 20 = 0$$

$$a=−2, −10$$

## Exercise $$\PageIndex{6}$$

$$b^2 - 4b + 4 = 0$$

## Exercise $$\PageIndex{7}$$

$$b^2 + 4b + 4 = 0$$

$$b=−2$$

## Exercise $$\PageIndex{8}$$

$$x^2 + 10x + 25 = 0$$

## Exercise $$\PageIndex{9}$$

$$2x^2 - 5x - 3 = 0$$

$$x = 3, -\dfrac{1}{2}$$

## Exercise $$\PageIndex{10}$$

$$6y^2 + y - 2 = 0$$

## Exercise $$\PageIndex{11}$$

$$4x^2 - 2x - 1 = 0$$

$$x = \dfrac{1 \pm \sqrt{5}}{4}$$

## Exercise $$\PageIndex{12}$$

$$3y^2 + 2y - 1 = 0$$

## Exercise $$\PageIndex{13}$$

$$5a^2 - 2a - 3 = 0$$

$$a = 1, -\dfrac{3}{5}$$

## Exercise $$\PageIndex{14}$$

$$x^2 - 3x + 1 = 0$$

## Exercise $$\PageIndex{15}$$

$$x^2 - 5x - 4 = 0$$

$$x = \dfrac{5 \pm \sqrt{41}}{2}$$

## Exercise $$\PageIndex{16}$$

$$(x+2)(x−1)=1$$

## Exercise $$\PageIndex{17}$$

$$(a+4)(a−5)=2$$

$$a = \dfrac{1 \pm \sqrt{89}}{2}$$

## Exercise $$\PageIndex{18}$$

$$(x−3)(x+3)=7$$

## Exercise $$\PageIndex{19}$$

$$(b−4)(b+4)=9$$

$$b = \pm 5$$

## Exercise $$\PageIndex{20}$$

$$x^2 + 8x = 2$$

## Exercise $$\PageIndex{21}$$

$$y^2 = -5y + 4$$

$$y = \dfrac{-5 \pm \sqrt{41}}{2}$$

## Exercise $$\PageIndex{22}$$

$$x^2 = -3x + 7$$

## Exercise $$\PageIndex{23}$$

$$x^2 = -2x - 1$$

$$x=−1$$

## Exercise $$\PageIndex{24}$$

$$x^2 + x + 1 = 0$$

## Exercise $$\PageIndex{25}$$

$$a^2 + 3a - 4 = 0$$

$$a=−4, 1$$

## Exercise $$\PageIndex{26}$$

$$y^2 + y = -4$$

## Exercise $$\PageIndex{27}$$

$$b^2 + 3b = -2$$

$$b=−1, −2$$

## Exercise $$\PageIndex{28}$$

$$x^2 + 6x + 8 = -x - 2$$

## Exercise $$\PageIndex{29}$$

$$x^2 + 4x = 2x - 5$$

No real number solution.

## Exercise $$\PageIndex{30}$$

$$6b^2 + 5b - 4 = b^2 + b + 1$$

## Exercise $$\PageIndex{31}$$

$$4a^2 + 7a - 2 = -2a + a$$

$$\dfrac{-2 \pm \sqrt{6}}{2}$$

## Exercise $$\PageIndex{32}$$

$$(2x + 5)(x - 4) = x^2 -x + 2$$

## Exercise $$\PageIndex{33}$$

$$(x-4)^2 = 3$$

$$x = 4 \pm \sqrt{3}$$

## Exercise $$\PageIndex{34}$$

$$(b - 6)^2 = 8$$

$$b = 6 \pm 2\sqrt{2}$$

## Exercise $$\PageIndex{35}$$

$$(3-x)^2 = 6$$

## Exercise $$\PageIndex{36}$$

$$3(x^2 + 1) = 2(x+7)$$

$$x = \dfrac{1 \pm \sqrt{34}}{3}$$

## Exercise $$\PageIndex{37}$$

$$2(y^2 - 3) = -3(y - 1)$$

## Exercise $$\PageIndex{38}$$

$$(x + 2)^2 = 4$$

## Exercise $$\PageIndex{39}$$

$$-4(a^2 + 2) + 3 = 5$$

No real number solution

## Exercise $$\PageIndex{40}$$

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

## Exercises For Review

Exercise $$\pageindex{41}$$.

Simplify $$(\dfrac{x^8y^7z^5}{x^4y^6z^2})^2$$

$$x^8y^2z^6$$

## Exercise $$\PageIndex{42}$$

Write $$4a^{-6}b^2c^3a^5b^{-3}$$ so that only positive exponents appear

## Exercise $$\PageIndex{43}$$

Find the product: $$(2y + 7)(3y - 1)$$

$$6y^2 + 19y - 7$$

## Exercise $$\PageIndex{44}$$

Simplify: $$\sqrt{80} - \sqrt{45}$$

## Exercise $$\PageIndex{45}$$

Solve $$x^2 - 4x - 12 = 0$$ by completing the square.

$$x=−2, 6$$

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## Course: Algebra 1   >   Unit 14

• Solve by completing the square: Integer solutions
• Solve by completing the square: Non-integer solutions
• Solve equations by completing the square
• Worked example: completing the square (leading coefficient ≠ 1)
• Completing the square
• Solving quadratics by completing the square: no solution
• Proof of the quadratic formula

## Solving quadratics by completing the square

• Completing the square review

## What you should be familiar with before taking this lesson

• Solving quadratic equations by taking the square root
• Solving quadratic equations by factoring

## What you will learn in this lesson

Solving quadratic equations by completing the square, what happened here, how to complete the square.

• The coefficient of x ‍   , which we know to be 6 ‍   , should be equal to 2 a ‍   . This means that a = 3 ‍   .
• The constant number we need to add is equal to a 2 ‍   , which is 3 2 = 9 ‍   .
• an integer, like 6 ‍
• a simplified proper fraction, like 3 / 5 ‍
• a simplified improper fraction, like 7 / 4 ‍
• a mixed number, like 1   3 / 4 ‍
• an exact decimal, like 0.75 ‍
• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍
• (Choice A)   b 2 ‍   A b 2 ‍
• (Choice B)   ( b 2 ) 2 ‍   B ( b 2 ) 2 ‍
• (Choice C)   ( 2 b ) 2 ‍   C ( 2 b ) 2 ‍
• (Choice D)   b 2 ‍   D b 2 ‍

## Solving equations one more time

• (Choice A)   x = 21 − 4 ‍   and − 21 − 4 ‍   A x = 21 − 4 ‍   and − 21 − 4 ‍
• (Choice B)   x = 69 − 8 ‍   and − 69 − 8 ‍   B x = 69 − 8 ‍   and − 69 − 8 ‍
• (Choice C)   x = 21 + 4 ‍   and − 21 + 4 ‍   C x = 21 + 4 ‍   and − 21 + 4 ‍
• (Choice D)   x = 69 + 8 ‍   and − 69 + 8 ‍   D x = 69 + 8 ‍   and − 69 + 8 ‍
• (Choice A)   x = 1 2 + 3 4 ‍   and − 1 2 + 3 4 ‍   A x = 1 2 + 3 4 ‍   and − 1 2 + 3 4 ‍
• (Choice B)   x = 2 + 3 2 ‍   and − 2 + 3 2 ‍   B x = 2 + 3 2 ‍   and − 2 + 3 2 ‍
• (Choice C)   x = 1 2 − 3 4 ‍   and − 1 2 − 3 4 ‍   C x = 1 2 − 3 4 ‍   and − 1 2 − 3 4 ‍
• (Choice D)   x = 2 − 3 2 ‍   and − 2 − 3 2 ‍   D x = 2 − 3 2 ‍   and − 2 − 3 2 ‍

## Arranging the equation before completing the square

Rule 1: separate the variable terms from the constant term, rule 2: make sure the coefficient of x 2 ‍   is equal to 1 ‍   ..

• (Choice A)   x = 7 − 5 2 ‍   and − 7 − 5 2 ‍   A x = 7 − 5 2 ‍   and − 7 − 5 2 ‍
• (Choice B)   x = 10 − 5 ‍   and − 10 − 5 ‍   B x = 10 − 5 ‍   and − 10 − 5 ‍
• (Choice C)   x = 7 + 5 2 ‍   and − 7 + 5 2 ‍   C x = 7 + 5 2 ‍   and − 7 + 5 2 ‍
• (Choice D)   x = 10 + 5 ‍   and − 10 + 5 ‍   D x = 10 + 5 ‍   and − 10 + 5 ‍

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Quadratic Formula Practice Problems with Answers. Below are ten (10) practice problems regarding the quadratic formula. The more you use the formula to solve quadratic equations, the more you become expert at it! ... Problem 9: Solve the quadratic equation using the quadratic formula. [latex]{\left( {2x - 1} \right)^2} = \Large{x \over 3 ...

Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. ... Number of solutions of quadratic equations. Quadratic formula review. Discriminant review. Math > Algebra 1 > ... quadratic formula (negative coefficients) Report a problem.

3. PDF Solve each equation with the quadratic formula.

Create your own worksheets like this one with Infinite Algebra 1. Free trial available at KutaSoftware.com. ©x d2Q0D1S2L RKcuptra2 GSRoYfRtDwWa8r9eb NLOL1Cs.j 4 lA0ll x TrCiagFhYtKsz OrVe4s4eTrTvXeZdy.c I RM8awd7e6 ywYiPtghR OItnLfpiqnAiutDeY QALlegpe6bSrIay V1g.N.

The Corbettmaths Practice Questions on the Quadratic Formula. Next: Rounding Significant Figures Practice Questions

5. 9.6: Solve Applications of Quadratic Equations

Step 5: Solve the equation. Substitute in the values. Distribute. This is a quadratic equation; rewrite it in standard form. Solve the equation using the Quadratic Formula. Identify the $$a,b,c$$ values. Write the Quadratic Formula. Then substitute in the values of $$a,b,c$$. Simplify. Figure 9.5.13: Rewrite to show two solutions.

Solve by completing the square: Non-integer solutions. Worked example: completing the square (leading coefficient ≠ 1) Solving quadratics by completing the square: no solution. Proof of the quadratic formula. Solving quadratics by completing the square. Completing the square review. Quadratic formula proof review.

Directions: Solve the following questions, for the indicated variable, using the quadratic formula. It is possible that some of these problems can also be solved by factoring, but for right now, we are practicing the quadratic formula. 1. Solve using the quadratic formula: x2 - 2 x - 24 = 0. Choose:

Solving quadratics by completing the square. Worked example: Completing the square (intro) Worked example: Rewriting expressions by completing the square. Worked example: Rewriting & solving equations by completing the square. Worked example: completing the square (leading coefficient ≠ 1) Solving quadratics by completing the square: no solution.

43. 7x2 − 9x = 0 7 x 2 − 9 x = 0. For the following exercises, solve the quadratic equation by using the square-root property. 44. x2 = 49 x 2 = 49. 45. (x − 4)2 = 36 ( x − 4) 2 = 36. For the following exercises, solve the quadratic equation by completing the square. 46. x2 + 8x − 5 = 0 x 2 + 8 x − 5 = 0. 47. 4x2 + 2x − 1 = 0 4 x ...

10. IXL

IXL plans. Virginia state standards. Textbooks. Test prep. Awards. Improve your math knowledge with free questions in "Solve a quadratic equation using the quadratic formula" and thousands of other math skills.

Previous: Factorising Quadratics Practice Questions Next: Adding Fractions Practice Questions GCSE Revision Cards

12. 9.5 Solve Applications of Quadratic Equations

This is a quadratic equation; rewrite it in standard form. Solve the equation using the Quadratic Formula. Identify the a, b, c a, b, c values. Write the Quadratic Formula. Then substitute in the values of a, b, c a, b, c. Simplify. Rewrite to show two solutions. Approximate the answers using a calculator. We eliminate the negative solution for ...

Quadratics by factoring. Solve for x . Enter the solutions from least to greatest. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

The Quadratic Formula. The Quadratic Formula: Given a quadratic equation in the following form: ax2 + bx + c = 0. ...where a, b, and c are the numerical coefficients of the terms of the quadratic, the value of the variable x is given by the following equation: \small { x = \dfrac {-b \pm \sqrt {b^2 - 4ac\phantom {\big|}}} {2a} } x= 2a−b± b2 ...

16. The Quadratic Formula to solve quadratic equations Step by step with

Example of the quadratic formula to solve an equation. Use the formula to solve theQuadratic Equation: y = x2 + 2x + 1 y = x 2 + 2 x + 1 . Just substitute a,b, and c into the general formula: a = 1 b = 2 c = 1 a = 1 b = 2 c = 1. Below is a picture representing the graph of y = x² + 2x + 1 and its solution.

The solutions to a quadratic equation of the form ax2 + bx + c = 0, a ≥ 0 are given by the formula: x = − b ± √b2 − 4ac 2a. To use the Quadratic Formula, we substitute the values of a, b, and c into the expression on the right side of the formula. Then, we do all the math to simplify the expression.

18. Solve quadratic equations by taking square roots (practice)

Quadratics by taking square roots. Google Classroom. You might need: Calculator. Solve for x . Enter the solutions from least to greatest. ( x + 5) 2 − 64 = 0. lesser x =. greater x =. Report a problem.

The solutions to the quadratic equation, as provided by the Quadratic Formula, are the x-intercepts of the corresponding graphed parabola. How? Well, when y = 0, you're on the x-axis. The x-intercepts of the graph are where the parabola crosses the x-axis. You're applying the Quadratic Formula to the equation ax 2 + bx + c = y, where y is set ...

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