### All Algebra II Resources

## Example Questions

### Example Question #244 : Equations / Inequalities

Solve the equation:

**Possible Answers:**

**Correct answer:**

Add 8 to both sides to set the equation equal to 0:

To factor, find two integers that multiply to 24 and add to 10. 4 and 6 satisfy both conditions. Thus, we can rewrite the quadratic of three terms as a quadratic of four terms, using the the two integers we just found to split the middle coefficient:

Then factor by grouping:

Set each factor equal to 0 and solve:

and

### Example Question #1 : Finding Zeros Of A Polynomial

Find the roots of the following quadratic expression:

**Possible Answers:**

**Correct answer:**

First, we have to know that "finding the roots" means "finding the values of x which make the expression =0." So basically we are going to set the original expression = 0 and factor.

This quadratic looks messy to factor by sight, so we'll use factoring by composition. We multiply a and c together, and look for factors that add to b.

So we can use 8 and -3. We will re-write 5x using these numbers as 8x - 3x, and then factor by grouping.

Note the extra + sign we inserted to make sure the meaning is not lost when parentheses are added. Now we identify common factors to be "pulled" out.

Now we factor out the (3x + 4).

Setting each factor = 0 we can find the solutions.

So the solutions are x = 1/2 and x = -4/3, or {-4/3, 1/2}.

### Example Question #4 : Solving Quadratic Functions

Find the roots of the following quadratic expression.

**Possible Answers:**

**Correct answer:**

First we remember that "find the roots" means "find the values of x for which this expression equals 0." So we set the expression = 0 and approach solving as normal.

Since solving this by sight is difficult, we'll use composition, multiplying a by c and finding factors which add to b.

So -9 and 5 will work; we will use them to rewrite -4x as -9x + 5x and then factor by grouping.

We identify common factors to "pull" out of each group.

And now we factor out x-3.

Setting each factor equal to 0 lets us solve for x.

So our solutions are x = -5/3 and x = 3, which we write as x = {-5/3, 3}.

### Example Question #11 : Finding Roots

FInd the roots for

**Possible Answers:**

**Correct answer:**

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 #12 : Finding Roots

Find the roots for

**Possible Answers:**

**Correct answer:**

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:**

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:**

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 .

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:**

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

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:**

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 .

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

### Example Question #17 : Finding Roots

Solve for :

**Possible Answers:**

**Correct answer:**

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

.

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