How to find the solution to a quadratic equation - Algebra 1

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

Factors of 10 include:

1 * 10= 10 1 + 10 = 11

2 * 5 =10 2 + 5 = 7

–2 * –5 = 10 –2 + –5 = –7 Good!

1. Now factor by grouping, pulling "x" out of the first pair and "-5" out of the second.

1. Now pull out the common factor, the "(x-2)," from both terms.

1. Set both terms equal to zero to find the possible roots and solve using inverse operations.

x – 5 = 0, x = 5

x – 2 = 0, x = 2

1. First step of solving any equation: combine like terms. With quadratics, the easiest step to take is to set the expression equal to zero.

1. There are two ways to do this problem. The first and most intuitive method is standard factoring.

16 + 1 = 17

8 + 2 = 10

4 + 4 = 8

1. Then follow the usual steps, pulling out the common factor from both pairs, "x" from the first and "4" from the second.

1. Pull out the "(x+4)" to wind up with:

1. Set each term equal to zero.

x + 4 = 0, x = –4

But there's a shortcut! Assuming the terms are arranged by descending degree (i.e., ), and the third term is both a perfect square whose square root is equal to half of the middle term, mathematicians use a little trick. In this case, the square root of 16 is 4. 4 * 2=8, so the trick will work. Take the square root of the first and last term, then stick a plus sign in between them and square the parentheses.

And x, once again, is equal to –4.

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. Begin the problem by factoring the final term. Include the negative when factoring.

–2 + 2 = 0

–4 + 1 = –3

–1 + 4 = 3

All options are exhausted, therefore the problem cannot be solved by factoring, which means that the roots either do not exist or are not rational numbers. We must use the quadratic formula.

1. Quadratics must be set equal to zero in order to be solved. To do so in this equation, the "8" has to wind up on the left side and combine with any other lone integers. So, multiply out the terms in order to make it possible for the "8" to be added to the other number.

Then combine like terms.

1. Now factor.

1 + 16 = 17

4 + 4 = 8

2 + 8 = 10

1. Pull out common factors, "x" and "8," respectively.

1. Pull out "(x+2)" from both terms.

x = –8, –2

1. Combine like terms and simplify.

No further simplification is possible. The first term has a coefficient that can't be factored away. FOIL requires that all terms be multiplied by each other at some point, so the presence of the coefficient has to be reflected in every step of the factoring.

1. Practically speaking, that means we add an extra step. Multiply the coefficient of the first term by the last term before factoring.

3 * 6 = 18

Factors of 18 include:

1 + 18 = 19

2 + 9 = 11

1. Now pull out the common factor in each of the pairs, "3x" from the first two and "2" from the second two.

1. Pull out the "(x+3)" from both terms.

1. Set both parts equal to zero and solve.

3x + 2 = 0, x = –2/3

x + 3 = 0, x = –3

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

1. Quadratics tend to be easier to follow when stated in order of descending degree. In other words, we need to rearrange the euqation.

1. No other simplification is possible, as there are no common factors between 15 and 4. Multiply the first coefficient by the final term and list off factors.

4 * –4 = –16

Factors of –16 include:

–1 + 16 = 15

1 + –16 = –15

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

1. Factor by pulling the greatest common factor out of each pair of terms, "x" from the first and "-4" from the second.

1. Factor out the "4x+1" from both terms.

1. Set both parts equal to zero and solve.

x – 4 = 0, x = 4

4x + 1 = 0, x = –1/4

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.

There are two ways to do this. One way involves using the quadratic formula. The quadratic formula is written below.

By looking at , a = 7, b = –4, and c = 13. Plug these values into the quadratic equation to find x.

Note that .

Factor out the two, then cancel out that two and separate terms.

This is our answer by the first merthod.

The other method to solve involves completing the square.

Subtract 13 to both sides.

Divide 7 to both sides.

Take the –4/7 from the x-term, cut it in half to get –2/7. Square that –2/7 to get 4/49. Finally, add 4/49 to both sides

Factor the left hand side and simplify the right hand side.

Square root and add 2/7 to both sides.

Don't forget to write it in terms of 'i'.

Note that we should find the same answer by either method.

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:

One can find the points of intersection of these two functions by setting them equal to one another, essentially substituting in one equation for the -side of the other equation. This will tell us when the (output) will be the same in each equation for a given (input).

By simplifying this equation and setting it equal to zero, we can find the two -values that produce the same values in the system of the two equations.

Subtract from both sides of the equation, and add to both sides.

Factoring this last equation makes it easier to find the -values that will result in zero on the left side of the equation. Set the two parenthetical phrases equal to zero to find two separate -values that satisfy the equation. These values will be the values of the points of intersection between the two equations.

We know our factors multiply to , and the six times one factor plus the other is equal to

and , so and are our factors.

Substituting these two values into either of the two original equations results in the -values of the points of intersection.

are the points of intersection.