Algebra II : Quadratic Equations and Inequalities

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #12 : Finding Roots

Find the roots for 

Possible Answers:

Correct answer:

Explanation:

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

 

The perfect square factors of this term are  and  .

Now we look at the constant term

The perfect square factors of this term are   and  

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

                    

                            

                              

                                                                

                                       

                                         

 

Example Question #13 : Finding Roots

Find the roots for 

Possible Answers:

Correct answer:

Explanation:

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

 

The perfect square factors of this term are  and  .

Now we look at the constant term

The perfect square factors of this term are  and 

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

                   

                                                          

                                       

                                         

 

Example Question #14 : Finding Roots

Find the roots for 

Possible Answers:

Correct answer:

Explanation:

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

 

The perfect square factors of this term are  and .

Now we look at the constant term

The perfect square factors of this term are  and

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

             

                     

                       

                               

                                       

                                         

 

Example Question #15 : Finding Roots

Find the roots of .

Possible Answers:

Correct answer:

Explanation:

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each term we see that each is a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

 

The perfect square factors of this term are  and .

Now we look at the constant term

The perfect square factors of this term are  and 

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

                               

                                       

                                         

Example Question #16 : Finding Roots

Find the roots of .

Possible Answers:

Correct answer:

Explanation:

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each term we see that each is a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

The perfect square factors of this term are  and .

Now we look at the constant term

The perfecct square factors of this term are  and .

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

                    

                            

                               

Example Question #12 : Finding Roots

Solve for :

Possible Answers:

Correct answer:

Explanation:

To solve the quadratic equation, , 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 #21 : Finding Roots

Solve for :

Possible Answers:

Correct answer:

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 #22 : Solving Quadratic Equations

Solve for :

Possible Answers:

Correct answer:

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 #2 : Finding Zeros Of A Polynomial

Find the roots of .

Possible Answers:

Correct answer:

Explanation:

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

 and 

 and 

Example Question #1481 : Algebra Ii

Find the roots of the quadratic expression .

Possible Answers:

Correct answer:

Explanation:

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

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