# Algebra 1 : Quadratic Equations

## Example Questions

### Example Question #41 : Quadratic Equations

Find the roots of the quadratic equation:

Explanation:

Reverse Foil Method

Factors of

Regroup

Roots occur at

### Example Question #461 : College Algebra

A farmer is building a fence around a field. He knows that the length of the field is 11 meters more than twice its width. If he knows that the area of the field is 30 square meters, what is the perimeter, in meters, of the field?

45

19

17

34

30

34

Explanation:

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 and solve for the perimeter.

Perimeter is found using the formula:

The perimeter of the field is 34 meters.

### Example Question #1 : How To Factor The Quadratic Equation

Find the Domain.

Explanation:

Given the equation

Realizing that the domain is restricted by the denominator, meaning that the denominator can not be equal to 0.

Set the denominator = to 0 and solve.

First factoring

Zero-product Property, setting both quantities equal to 0 and solve.

So when x is 6 or -1, our denominator will be 0. Meaning those would be our domain restrictions.

### Example Question #1 : How To Factor The Quadratic Equation

Find the roots of the equation x2 + 5x + 6 = 0

2 and 3

–5 and 1

1 and –3

–2 and –3

3 and –3

–2 and –3

Explanation:

To factor this, we need to find a pair of numbers that multiplies to 6 and sums to 5. The numbers 2 and 3 work. (2 * 3 = 6 and 2 + 3 = 5)

So (x + 2)(x + 3) = 0

x = –2 or x = –3

### Example Question #1 : How To Factor The Quadratic Equation

Find the domain:

Explanation:

To find the domain, find all areas of the number line where the fraction is defined.

because the denominator of a fraction must be nonzero.

Factor by finding two numbers that sum to -2 and multiply to 1.  These numbers are -1 and -1.

### Example Question #2351 : Algebra 1

Find the solutions to the equation .

No solution

Explanation:

To factor the polynomial, we need the numbers that multiply to –12 and add to +1. This leads us to –3 and +4. We solve the polynomial by setting it equal to 0.

So either x = 3 or x = –4 will make the product equal to 0.

### Example Question #341 : Intermediate Single Variable Algebra

Solve for x.

Explanation:

This is a factoring problem, so we need to get all of the variables on one side and set the equation equal to zero. To do this we subtract 128 from both sides to get .

We then notice that all four numbers are divisible by four, so we can simplify the expression to .

Think of the equation in this format to help with the following explanation.

We must then factor to find the solutions for x. To do this we must make a factor tree of c (which is 32 in this case) to find the possible solutions. The possible numbers are 1 * 32, 2 * 16, and 4 * 8.

Since c is negative, we know that our factoring will produce a positive and negative number.

We then look at b to see if the greater number will be positive or negative. Since b is positive, we know that the greater number from our factoring tree will be positive.

We then use addition and subtraction with the factoring tree to find the numbers that add together to equal b. Remember that the greater number is positive and the lesser number is negative in this example.

Positive 8 and negative 4 equal b. We then plug our numbers into the factored form of .

We know that anything multiplied by 0 is equal to 0, so we plug in the numbers for x which make each equation equal to 0. In this case .

### Example Question #2352 : Algebra 1

Find solutions to .

Explanation:

The quadratic can be solved as . Setting each factor to zero yields the answers.

Explanation:

### Example Question #1 : How To Factor The Quadratic Equation

Explanation:

When we attempt to factor a quadratic, we must first look for the factored numbers. When quadratics are expressed as  the factored numbers are  and . Since , we know the factors for 1 are 1 and 1. So we know the terms will be

Looking at our constant, , we see a positive 6. So 6 factors into either 2 and 3  or 1 and 6 (since  and ). Since our constant is a positive number, we know that our factors are either both positive, or both negative. (Note: you should know that 2 negative numbers multiplied becomes a positive number).

So to figure out what we must use we look at the  part of the quadratic. We are looking for 2 numbers which add up to our . So, 1 and 6 do not work, since . But, 2 and 3 are perfect since .

But, since our  is a negative 5, we know we must use negative numbers in our factored expression. Thus, our factoring must become

or