### All ACT Math Resources

## Example Questions

### Example Question #181 : Algebra

If then which of the following is a possible value for ?

**Possible Answers:**

**Correct answer:**

Since , .

Thus

Of these two, only 4 is a possible answer.

### Example Question #41 : Quadratic Equations

Find all real solutions to the equation.

**Possible Answers:**

**Correct answer:**

To solve by factoring, we need two numbers that add to and multiply to .

In order for the equation to equal zero, one of the terms must be equal to zero.

OR

Our final answer is that .

### Example Question #21 : Quadratic Equation

How many real solutions are there for the following equation?

**Possible Answers:**

**Correct answer:**

The first thing to notice is that you have powers with a regular sequence. This means you can simply treat it like a quadratic equation. You are then able to factor it as follows:

The factoring can quickly be done by noticing that the 14 must be either or . Because it is negative, one constant will be negative and the other positive. Finally, since the difference between 14 and 1 cannot be 5, it must be 7 and 2.

Alternatively, one could use the quadratic formula.

The end result is that you have:

The latter of these two gives only complex answers, so there are two real answers.

### Example Question #301 : Equations / Inequalities

The formula to solve a quadratic expression is:

All of the following equations have real solutions EXCEPT:

**Possible Answers:**

**Correct answer:**

We can use the quadratic formula to find the solutions to quadratic equations in the form ax^{2 }+ bx + c = 0. The quadratic formula is given below.

In order for the formula to give us real solutions, the value under the square root, b^{2 }– 4ac, must be greater than or equal to zero. Otherwise, the formula will require us to find the square root of a negative number, which gives an imaginary (non-real) result.

In other words, we need to look at each equation and determine the value of b^{2 }–^{ }4ac. If the value of b^{2 }– 4ac is negative, then this equation will not have real solutions.

Let's look at the equation 2x^{2} – 4x + 5 = 0 and determine the value of b^{2 }– 4ac.

b^{2 }– 4ac = (–4)^{2} – 4(2)(5) = 16 – 40 = –24 < 0

Because the value of b^{2 }– 4ac is less than zero, this equation will not have real solutions. Our answer will be 2x^{2} – 4x + 5 = 0.

If we inspect all of the other answer choices, we will find positive values for b^{2 }– 4ac, and thus these other equations will have real solutions.

### Example Question #41 : Quadratic Equations

Let , and let . What is the sum of the possible values of such that .

**Possible Answers:**

**Correct answer:**

We are told that f(x) = x^{2} - 4x + 2, and g(x) = 6 - x. Let's find expressions for f(k) and g(k).

f(k) = k^{2} – 4k + 2

g(k) = 6 – k

Now, we can set these expressions equal.

f(k) = g(k)

k^{2} – 4k +2 = 6 – k

Add k to both sides.

k^{2} – 3k + 2 = 6

Then subtract 6 from both sides.

k^{2} – 3k – 4 = 0

Factor the quadratic equation. We must think of two numbers that multiply to give us -4 and that add to give us -3. These two numbers are –4 and 1.

(k – 4)(k + 1) = 0

Now we set each factor equal to 0 and solve for k.

k – 4 = 0

k = 4

k + 1 = 0

k = –1

The two possible values of k are -1 and 4. The question asks us to find their sum, which is 3.

The answer is 3.

### Example Question #581 : Algebra

What are the roots of the quadratic equation given by:

**Possible Answers:**

**Correct answer:**

To find the roots of a quadratic equation, begin by factoring the quadratic. To do this, factor the first term () and the final term (in our case ). This gives us (remember, the two numbers you choose should add to the middle term, and multiple to the final term. and )

Then, the roots are the values that make the equation zero. So set each parentheses equal to zero and solve for .

### Example Question #581 : Algebra

Solve for :

Round to the nearest hundredth.

**Possible Answers:**

**Correct answer:**

With quadratic equations, you should *always* start by getting all of your terms to one side of the equation, setting this equal to :

Thus, simplify into:

Now, the next question you need to ask yourself is, "Can this be factored?" In this case, it can. Factor the quadratic expression:

Now, remember that you merely need to set each group equal to . This gives you the two values for :

; therefore

Likewise, for the other group,

### Example Question #51 : Quadratic Equations

Solve for :

**Possible Answers:**

**Correct answer:**

With quadratic equations, you should *always* start by getting all of your terms to one side of the equation, setting this equal to :

Thus, simplify into:

Now, the next question you need to ask yourself is, "Can this be factored?" In this case, it can, though we are sometimes a bit intimidated by terms that have a coefficient like this. Factor the quadratic expression:

If you FOIL this out, you will see your original equation.

Now, remember that you merely need to set each group equal to . This gives you the two values for :

For the other group, you get .

### Example Question #41 : How To Find The Solution To A Quadratic Equation

Solve for :

**Possible Answers:**

**Correct answer:**

With quadratic equations, you should *always* start by getting all of your terms to one side of the equation, setting this equal to :

Thus, simplify into:

Now, the next question you need to ask yourself is, "Can this be factored?" In this case, it cannot be easily factored. Therefore, you should use the quadratic formula. Recall that its general form is:

For our data, , , and .

Thus, we have:

Simplifying, this is:

Now, simplify.

### Example Question #53 : Quadratic Equations

Solve for :

**Possible Answers:**

**Correct answer:**

*always* start by getting all of your terms to one side of the equation, setting this equal to :

Thus, simplify into:

Now, the next question you need to ask yourself is, "Can this be factored?" In this case, it cannot be easily factored. Therefore, you should use the quadratic formula. Recall that its general form is:

For our data, , , and .

Thus, we have:

Simplifying, this is:

Since is negative, you know that there is no real solution (given the problems arising from the negative square root)!