Quadratic Equations - College Algebra

Write the equation in standard form by first eliminating parentheses, then moving all terms to the left of the equal sign.

First:

Inside:

Outside:

Last:

Now factor, set each binomial to zero, and solve individually. We are lookig for two numbers with sum and product ; these numbers are .

and

or

The solution set is .

Write the equation in standard form by first eliminating parentheses, then moving all terms to the left of the equal sign.

First:

Inside:

Outside:

Last:

Now factor, set each binomial to zero, and solve individually. We are lookig for two numbers with sum and product ; these numbers are .

and

or

The solution set is .

Write the equation in standard form by first eliminating parentheses, then moving all terms to the left of the equal sign.

First:

Inside:

Outside:

Last:

Now factor, set each binomial to zero, and solve individually. We are lookig for two numbers with sum and product ; these numbers are .

and

or

The solution set is .

1. This is a relatively standard quadratic equation. List and add factors of 18.

1 + 18 = 19

2 + 9 = 11

3 + 6 = 9

1. Pull out common factors of each pair, "x" from the first and "6" from the second.

1. Factor again, pulling out "(x+3)" from both terms.

1. Set each term equal to zero and solve.

x + 3 = 0, x = –3

x + 6 = 0, x = –6

1. After adding like terms and setting the equation equal to zero, the immediate next step in solving any quadratic is to simplify. If the coefficients of all three terms have a common factor, pull it out. So go ahead and divide both sides (and therefore ALL terms on BOTH sides) by 4.

Since zero divided by four is still zero, only the left side of the equation changes.

1. Either factor by grouping or use the square trick.

Grouping:

1 + 1 = 2

(The "1" was pulled out only to make the next factoring step clear.)

x + 1 = 0, x = –1

OR

Perfect Square:

x = –1

1. The first step would be to simplify, but since 2, 15, and 25 have no common factors greater than 1, simplification is impossible.

Now we factor. Multiply the first coefficient by the final term and list off the factors.

2 * 25 = 50

Factors of 50 include:

1 + 50 = 51

2 + 25 = 27

5 + 10 = 15

1. Split up the middle term to make factoring by grouping possible.

Note that the "2" and the "10," and the "5" and the "25," have to go together for factoring to come out with integers. Always make sure the groups actually have a common factor to pull.

1. Pull out the common factors from both groups, "2x" from the first and "5" from the second.

1. Factor out the "(x+5)" from both terms.

1. Set each parenthetical expression equal to zero and solve.

2x + 5 = 0, x = –5/2

x + 5 = 0, x = –5

In order to find the perimeter, start by defining the variables. It is typically easier to define one of the variables in terms of the other; therefore, only one unknown will need to be calculated to find the perimeter. The problem states that the length is eleven more than twice the width; thus, we can define our variables in the following way:

The farmer knows that the field's area is thirty square meters. Area is found using the following formula:

Substitute in the known value for the area and the defined variables for the length and width.

Notice that this equation possesses all the components of a quadratic. Use the information in the equation to construct a quadratic equation that can be factored to obtain an answer. Start by multiplying the first term by the variable on the right side of the equation.

In order to make the quadratic equal to zero, subtract 30 from both sides of the equation.

Now, factor the quadratic and solve for the variable. We can use the ac method to solve for the variable. Quadratics can be written in the following format:

We need to find two numbers whose product equals a multiplied by c and whose sum equals b; therefore, the product of the factors must be -60 and their sum must equal 11. Write out the prime factorization of 60.

There is one factor of -60 that when added together sum to equal 11: 15 and -4.

Use the factors and split the middle term in the quadratic in order to make factoring by grouping possible.

Pull the greatest common factor from each pair of terms: from the first and 15 from the second.

Factor out the quantity from both terms.

Set each factor equal to zero and solve for w.

We can cross out the this negative option because the width of a dimension cannot be a negative value. Solve for w in the second factor.

The width of the field is 2 meters. Substitute 2 in for the variable w and solve for the perimeter.

Perimeter is found using the formula:

The perimeter of the field is 34 meters.

1. Before we can figure out when Billy will be 1.5 times Johnny's age, we have to figure out their current ages. So let's define our variables in terms of the first part of the question.

B = Billy's age and J = Johnny's age

It's easier to solve if we put one variable in terms of the other. If Billy were just twice as old as Johnny, we could write his age as B = 2J.

But Billy is one less than twice as old as Johnny, so B = 2J – 1

1. We know that the two boys' ages multiply together to make ninety-one.

B * J = J(2J – 1) = 91

1. Now we have our factored quadratic. We just need to multiply it out and set everything equal to zero to begin.

1. Now we need to factor back out. We start by multiplying the first coefficient by the final term and listing off the factors.

2 * –91 = –182

1 + –182 = –181

2 + –91 = –89

7 + –26 = –19

13 + –14 = –1

1. Split up the middle term in so that factoring by grouping is possible.

1. Factor by grouping, pulling out "2J" from the first set of terms and "13" from the second.

1. Factor out the "(J-7)" from both terms.

1. Set both parentheses equal to zero and solve.

2J + 13 = 0, J = –13/2

J – 7 = 0, J = 7

Clearly, only of the two solutions works, since Johnny's age has to be positive. Johnny is 7, therefore Billy is 2(7) – 1=13. But we're not done yet!

1. We need to figure out at what point Billy will 1.5 times Johnny's age. Guess and check would be a fairly efficient way to do this problem, but setting up an equation would be even faster. First, though, we need to figure out what our variable is. We know Billy's and Johnny's current ages; we just need to figure out their future ages. One variable is always better than two, so instead of using two different variables to represent their respective future ages, we'll use one variable to represent the number of years we have to add to each of their current ages in order to make Billy 1.5 times older than Johnny. Let's call that variable "x."

1.5(J + x) = B + x

We know the values of J and B, so we can go ahead and fill those in.

1.5(7 + x) = 13 + x

1. Then we solve for x algebraically, with inverse order of operations.

10.5 + 1.5x = 13 + x

0.5x = 2.5

x = 5

J = 7 + 5 = 12

B = 13 + 5 = 18

This requires the use of the quadratic formula. Recall that:

for .

For this problem, .

So,

.

.

Therefore, the two solutions are:

First, we must factor out any common factors between the two terms. Both 3 and 12 share a factor of 3, so we can "take" 3 out, like this:

.

Inside the parentheses, it becomes clear that this is a difference of squares problem (a special factor), which can be solved with the equation

.

Thus, .

Now, we can set each factor to 0, and solve for :

and

.

Solve by factoring. We need to find two factors that multiply to eight and add to six.

One of these factors must equal zero in order for the equation to be true.

The quadratic formula is as follows:

We will start by finding the values of the coefficients of the given equation:

Quadratic equations may be written in the following format:

In our case, the values of the coefficients are:

Substitute the coefficient values into the quadratic equation:

After simplifying we are left with:

leaving us with our two solutions:

and

To solve for x, we must first simplify the trinomial into two binomials.

To simplify the trinomial, its general form given by , we must find factors of that when added give us . For our trinomial, and the two factors that add together to get are and .

Now, using the two factors, we can rewrite as the sum of the two factors each multiplied by x:

Next, we group the first two and last two terms together and factor each of the groups:

Now, simplify further:

Finally, set each of these binomials equal to zero and solve for x:

To solve for , we can use the quadratic formula:

Add 3 to both sides of the equation, to isolate the x terms:

Divide each term by 8, to isolate the x2 term:

Add the square of one half of the "b" term to each side:

Simplify:

Simplify further:

Use the following factoring rule to simplify the left side of the equation:

Take the square root of both sides of the equation:

Simplify:

Subtract one eighth from each side of the equation: or

Simplify the equations:

Solution:

1. This is a relatively standard quadratic equation. List and add factors of 18.

1 + 18 = 19

2 + 9 = 11

3 + 6 = 9

1. Pull out common factors of each pair, "x" from the first and "6" from the second.

1. Factor again, pulling out "(x+3)" from both terms.

1. Set each term equal to zero and solve.

x + 3 = 0, x = –3

x + 6 = 0, x = –6

1. After adding like terms and setting the equation equal to zero, the immediate next step in solving any quadratic is to simplify. If the coefficients of all three terms have a common factor, pull it out. So go ahead and divide both sides (and therefore ALL terms on BOTH sides) by 4.

Since zero divided by four is still zero, only the left side of the equation changes.

1. Either factor by grouping or use the square trick.

Grouping:

1 + 1 = 2

(The "1" was pulled out only to make the next factoring step clear.)

x + 1 = 0, x = –1

OR

Perfect Square:

x = –1

1. The first step would be to simplify, but since 2, 15, and 25 have no common factors greater than 1, simplification is impossible.

Now we factor. Multiply the first coefficient by the final term and list off the factors.

2 * 25 = 50

Factors of 50 include:

1 + 50 = 51

2 + 25 = 27

5 + 10 = 15

1. Split up the middle term to make factoring by grouping possible.

Note that the "2" and the "10," and the "5" and the "25," have to go together for factoring to come out with integers. Always make sure the groups actually have a common factor to pull.

1. Pull out the common factors from both groups, "2x" from the first and "5" from the second.

1. Factor out the "(x+5)" from both terms.

1. Set each parenthetical expression equal to zero and solve.

2x + 5 = 0, x = –5/2

x + 5 = 0, x = –5

In order to find the perimeter, start by defining the variables. It is typically easier to define one of the variables in terms of the other; therefore, only one unknown will need to be calculated to find the perimeter. The problem states that the length is eleven more than twice the width; thus, we can define our variables in the following way:

The farmer knows that the field's area is thirty square meters. Area is found using the following formula:

Substitute in the known value for the area and the defined variables for the length and width.

Notice that this equation possesses all the components of a quadratic. Use the information in the equation to construct a quadratic equation that can be factored to obtain an answer. Start by multiplying the first term by the variable on the right side of the equation.

In order to make the quadratic equal to zero, subtract 30 from both sides of the equation.

Now, factor the quadratic and solve for the variable. We can use the ac method to solve for the variable. Quadratics can be written in the following format:

We need to find two numbers whose product equals a multiplied by c and whose sum equals b; therefore, the product of the factors must be -60 and their sum must equal 11. Write out the prime factorization of 60.

There is one factor of -60 that when added together sum to equal 11: 15 and -4.

Use the factors and split the middle term in the quadratic in order to make factoring by grouping possible.

Pull the greatest common factor from each pair of terms: from the first and 15 from the second.

Factor out the quantity from both terms.

Set each factor equal to zero and solve for w.

We can cross out the this negative option because the width of a dimension cannot be a negative value. Solve for w in the second factor.

The width of the field is 2 meters. Substitute 2 in for the variable w and solve for the perimeter.

Perimeter is found using the formula:

The perimeter of the field is 34 meters.