# Algebra II : Finding Roots

## Example Questions

### Example Question #21 : Solving Quadratic Equations

Solve for :

Explanation:

To solve the quadratic equation,  , let's first set the equation equal to zero and then factor the quadratic to

.

Since the equation is equal to zero, at least one of the expressions on the left side of the equation must be equal to zero. Therefore, set  and solve to get

.

Finally, set  to get

.

The solution to the original equation is

.

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

Solve for :

Explanation:

In order to solve the equation  , first set the expression equal to zero:

Then factor:

Since the equation is equal to zero, at least one of the expressions on the lefthand side of the equation must therefore be equal to zero. Starting with the first expression:

Finally, set the second expression to zero and solve for :

The solution to the equation is

.

### Example Question #342 : Intermediate Single Variable Algebra

Find the roots of .

Explanation:

If we recognize this as an expression with form , with  and , we can solve this equation by factoring:

and

and

### Example Question #343 : Intermediate Single Variable Algebra

Find the roots of the quadratic expression .

Explanation:

Looking at this expression, we can see it is of the form , with , and . Therefore, we can write it in the form :

### Example Question #344 : Intermediate Single Variable Algebra

Find the roots of the following equation:

Explanation:

A polynominal with the form  multiplies out to , so this leaves two equations:

Factors of 12 are 2, 6, and 3, 4. In this case, because the product is negative, one root is positive and one is negative. Because the sum is negative, the positive root must have a lower absolute value than the negative root. This leaves us with two possibilities:  2, -6 and 3, -4. Plugging into the sum equation shows the roots to be 3, -4. A check into the original polynominal shows

### Example Question #345 : Intermediate Single Variable Algebra

Solve for .

Explanation:

When asked to solve for , we are really searching for the roots/-intercepts of the equation.

In this particular case, our function is already factored for us leaving us with only a few steps to complete the problem.

Our first step is to set each term equal to , leaving us with...

and

The next step is to use our knowledge of order of operations to simply solve for  for each of the above equations...

Subtract  from both sides

Divide by

Answer # 2, our second root/-intercept

### Example Question #346 : Intermediate Single Variable Algebra

Solve for .

Explanation:

This problem requires simplification, order of operations, and knowledge of square roots.

Our goal is to isolate/solve for

Divide by  on both sides

Square root both sides

**Remember: DO NOT FORGET THAT WHEN WE TAKE A SQUARE ROOT, WE GET A PLUS/MINUS ANSWER.

or

and

Explanation:

.

### Example Question #1481 : Algebra Ii

Give all real solutions of the following equation:

The equation has no real solutions.

Explanation:

By substituting  - and, subsequently,  this can be rewritten as a quadratic equation, and solved as such:

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product 36 and sum ; these integers are .

Substitute back:

These factors can themselves be factored as the difference of squares:

Set each factor to zero and solve:

The solution set is .

### Example Question #21 : Finding Roots

Solve for x.

x = –6, –4

x = 6, 4

Cannot be factored by grouping.

x = 4

x = –8, –2

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