### All Algebra 1 Resources

## Example Questions

### Example Question #202 : Equations / Inequalities

Solve for x.

**Possible Answers:**

x = –2/3, –3

x = –5

x = –5/2, –5

x = –2/5, –5

x = –5, 5

**Correct answer:**

x = –5/2, –5

1) The first step would be to simplify, but since 2, 15, and 25 have no common factors greater than 1, simplification is impossible.

Now we factor. Multiply the first coefficient by the final term and list off the factors.

2 * 25 = 50

Factors of 50 include:

1 + 50 = 51

2 + 25 = 27

5 + 10 = 15

2) Split up the middle term to make factoring by grouping possible.

Note that the "2" and the "10," and the "5" and the "25," have to go together for factoring to come out with integers. Always make sure the groups actually have a common factor to pull.

3) Pull out the common factors from both groups, "2x" from the first and "5" from the second.

4) Factor out the "(x+5)" from both terms.

5) Set each parenthetical expression equal to zero and solve.

2x + 5 = 0, x = –5/2

x + 5 = 0, x = –5

### Example Question #1 : Quadratic Equations

Use the quadratic formula to find the solutions to the equation.

**Possible Answers:**

and

and

and

and

and

**Correct answer:**

and

The quadratic formula is as follows:

We will start by finding the values of the coefficients of the given equation:

Quadratic equations may be written in the following format:

In our case, the values of the coefficients are:

Substitute the coefficient values into the quadratic equation:

After simplifying we are left with:

leaving us with our two solutions:

and

### Example Question #31 : Quadratic Equations

Find the roots of the quadratic equation.

**Possible Answers:**

**Correct answer:**

Use the quadratic equation:

and use the general formula

to find the coefficients.

In our case, thus, , , .

Plugging our values into the quadratic equation we are able to solve for .

### Example Question #31 : How To Find The Solution To A Quadratic Equation

What are the zeroes of the equation ?

**Possible Answers:**

**Correct answer:**

Our first step is to factor the expression. We must find two numbers which have a sum of -4 and a product of -32. -32 can be factored into +/-2 and -/+16, or +/-4 and -/+8. The only combination that satisfies the product *and *sum is +4 and -8.

This means we can factor our expression to .

To find the zeroes, we simply need to find where our expression *equals* zero. And we can find this easily by determining where (x+4) and (x-8) each equal zero.

The zeroes of our expression, then, are at -4 and 8.

and

### Example Question #82 : Systems Of Equations

Solve for .

**Possible Answers:**

**Correct answer:**

1) Begin the problem by factoring the final term. Include the negative when factoring.

–2 + 2 = 0

–4 + 1 = –3

–1 + 4 = 3

All options are exhausted, therefore the problem cannot be solved by factoring, which means that the roots either do not exist or are not rational numbers. We must use the quadratic formula.

### Example Question #32 : How To Find The Solution To A Quadratic Equation

Find the solutions to this quadratic equation:

**Possible Answers:**

None of the other answers.

**Correct answer:**

Put the quadratic in standard form:

Factor:

An easy way to factor (and do so with less trial and error) is to think of what two numbers could multiply to equal "c", but add to equal "b". These letters come from the designations in the standard form of a quadratic equation: . As you can see the product of -6 and 2 is -12 and they both add to 4.

### Example Question #31 : How To Find The Solution To A Quadratic Equation

Find the roots of the following equation.

**Possible Answers:**

**Correct answer:**

Use the quadratic formula to solve the equation.

Plug in these values and solve.

### Example Question #31 : Quadratic Equations

Solve for :

**Possible Answers:**

or

None of these answers

or

or

**Correct answer:**

or

We start by subtracting on both sides in order to get a quadratic expression on one side of the equation:

We can factor this into:

Thus, either or .

So, or

### Example Question #1 : Quadratic Equations

Solve for x:

**Possible Answers:**

**Correct answer:**

To solve for x, we must first simplify the trinomial into two binomials.

To simplify the trinomial, its general form given by , we must find factors of that when added give us . For our trinomial, and the two factors that add together to get are and .

Now, using the two factors, we can rewrite as the sum of the two factors each multiplied by x:

Next, we group the first two and last two terms together and factor each of the groups:

Now, simplify further:

Finally, set each of these binomials equal to zero and solve for x:

### Example Question #38 : Quadratic Equations

Use the quadratic equation to determine the real roots for:

**Possible Answers:**

**Correct answer:**

Write the quadratic equation.

The quadratic is in the form:

Substitute the known coefficients.

Simplify the numerator and denominator.

Because we have a negative value inside the square root, we will have complex roots instead of real roots. The roots are imaginary.

The answer is:

Certified Tutor