# Algebra 1 : How to find the solution to a quadratic equation

## Example Questions

Explanation:

Explanation:

Explanation:

### Example Question #1 : Quadratic Equations

Solve for .

Explanation:

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.

### Example Question #11 : How To Find The Solution To A Quadratic Equation

Solve

Explanation:

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

.

### Example Question #61 : Systems Of Equations

The product between an integer and another integer that's  units greater is equal to . Find these integers.

none of these

Explanation:

We'll call one of the integers , which makes the other integer . The product between  and  is .  This is written as

Solve for .

We can solve this either by using the quadratic formula or by factoring.  Factoring seems to be the quickest method.

Take the square root of both sides.

The first integer is , which we can use to find the other integer by using .

The second integer is 2.

Therefore, the two integers are and .

### Example Question #11 : Quadratic Equations

Find the solutions:

Explanation:

Since the trinomial starts with , the first term in each of the binomials will be :

Since you have a  at the end of the trinomial, the two terms at the end of the two binomials must multiply to . Since it is a negative , the signs in the two binomials will be opposite.

and  work:

Now you set each binomial equal to  independently, and solve for .

### Example Question #192 : Equations / Inequalities

Explanation:

FOIL (first, outside, inside, last) is a device to help students remember to multiply every term by every other term exactly one time. The steps of using FOIL progress as follows:

1) Multiply the first term of each parenthetical expression together:

2) Next, multiply the "outside" terms together:

Add that to the first product, since the terms are all being added together in the parentheses, and the distributive property requires that we maintain the sign:

3) Multiply the "inside" terms together:

Add that to the first two products:

4) Finally, multiply the last terms from each parenthetical expression together:

Add all of the terms together and combine like terms where possible:

5) We now have a quadratic equation. In order to solve a quadratic, we must first set our expression equal to zero by subtracting 6 from both sides.

6) Next, we factor by using backwards distribution. The idea is to split up the middle term, , in such a way that one part has a common factor with  and the other part has a common factor with 6. We can then factor by grouping. Obviously, no matter how we split up , one part will always contain an . So we must focus our attention on the 6. What are the factors of 6 that add up to 7?

Yes!

So,

7) Now we pull out the common factor from each pair of terms,  from the first and 6 from the second:

8) Now we factor again, taking out the common factor  from each term:

9) We know that at least one of the terms must be equal to zero in order for the product to equal zero, so we set each part equal to zero and solve:

and

### Example Question #11 : How To Find The Solution To A Quadratic Equation

Where do the graphs of these two equations intersect?

Equation 1:

Equation 2:

Explanation:

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.

### Example Question #2 : Finding Roots

Solve for x.

x = –5, –2

x = –4, –3

x = 5

x = 5, 2

x = 4, 3

x = 5, 2

Explanation:

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!

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

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

4) 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