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

## Example Questions

### Example Question #1 : Finding Roots

Solve for x.

x = 5, 2

x = –4, 4

x = –5, –2

x = –4

x = 2

x = –4

Explanation:

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.

2) 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

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

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

5) 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.

### Example Question #1 : Finding Roots

Solve for :

Explanation:

To factor, find two numbers that sum to 5 and multiply to 6.

Check the possible factors of 6:

1 * 6 = 6

1 + 6 = 7, so these don't work.

2 * 3 = 6

2 + 3 = 5, so these work!

Next, pull out the common factors of the first two terms and then the second two terms:

Set both expressions equal to 0 and solve:

and

### Example Question #21 : How To Find The Solution To A Quadratic Equation

Solve for .

Explanation:

Notice that both terms have in them. Factor this out:

Set both factors equal to zero and solve for .

### Example Question #2321 : Algebra 1

Solve for

Explanation:

First, we want to use the quadratic formula for this equation, to find the values for x:

When the quadratic formula is simplified, it should look like this:

This provides us with 2 possible values for x:

and/or

To confirm whether or not these are values of X, we need to plug them in, to see if the equation makes sense:

12+24-36 = 0, x = 2

108-72-36 = 0, x = -6

Therefore,

x = 2 and x = -6

### Example Question #71 : Systems Of Equations

Solve for :

Explanation:

We know that this equation can be simplified to the following format:

(X+_)(X+_) = 0

X will be cross-multiplied by X to form the  of the equation. We need to plug in number values that will result in 4x and 3 when cross-multiplied:

(X+3) = 0, X = -3

(X+1) = 0, X = -1

### Example Question #21 : How To Find The Solution To A Quadratic Equation

Solve for :

Explanation:

When looking at the equation, it is clear that both coefficients and the number value are divisible by 5:

So our first step is to divide both sides of the equation by 5:

When further simplified, we get:

When we plug in the X-Value, we can confirm whether or not this is the correct value for X:

Thus we can confirm that this is the correct value for X.

### Example Question #21 : How To Find The Solution To A Quadratic Equation

Solve for  using the Quadratic Formula:

Explanation:

The first step to this quadratic equation is to divide it by the least common denominator, which in this case is 2:

Next, we plug the equation into the Quadratic Formula:

and/or

and/or

We now have 2 possible values for X. The next step is to plug both values in to determine if they are correct values of X:

After confirming both values for X work when plugged into the quadratic formula, we can confirm:

and

### Example Question #21 : How To Find The Solution To A Quadratic Equation

Solve for :

Explanation:

Since this is a quadratic formula and X-squared has a coefficient of 1, the simplified equation will be in the following format:

Looking at the simplified equation and the quadratic equation above, we can see that the numerical values we plug in need to sum up to 6X and have a product of -16 when cross-multiplied.

If we plug in 8 and -2, we will get the original equation when we cross-multiply:

To double check our answers, we can plug them in to make sure they make sense:

### Example Question #21 : Finding Roots

Solve for x.

x = –6, –4

x = –8, –2

Cannot be factored by grouping.

x = 4

x = 6, 4

x = –8, –2

Explanation:

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.

2) Now factor.

1 + 16 = 17

4 + 4 = 8

2 + 8 = 10

3) Pull out common factors, "x" and "8," respectively.

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

x = –8, –2

### Example Question #81 : Systems Of Equations

Solve for x.

x = –2/3, –3

x = –12, –1

x = 4, 1

x = –4, –1

x = –9, –2

x = –2/3, –3

Explanation:

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.

2) 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

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

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

5) Set both parts equal to zero and solve.

3x + 2 = 0, x = –2/3

x + 3 = 0, x = –3