Since , .

Thus

Of these two, only 4 is a possible 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 .

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.

We can use the quadratic formula to find the solutions to quadratic equations in the form ax² + 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, b² – 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 b² – 4ac. If the value of b² – 4ac is negative, then this equation will not have real solutions.

Let's look at the equation 2x² – 4x + 5 = 0 and determine the value of b² – 4ac.

b² – 4ac = (–4)² – 4(2)(5) = 16 – 40 = –24 < 0

Because the value of b² – 4ac is less than zero, this equation will not have real solutions. Our answer will be 2x² – 4x + 5 = 0.

If we inspect all of the other answer choices, we will find positive values for b² – 4ac, and thus these other equations will have real solutions.

We are told that f(x) = x² - 4x + 2, and g(x) = 6 - x. Let's find expressions for f(k) and g(k).

f(k) = k² – 4k + 2

g(k) = 6 – k

Now, we can set these expressions equal.

f(k) = g(k)

k² – 4k +2 = 6 – k

Add k to both sides.

k² – 3k + 2 = 6

Then subtract 6 from both sides.

k² – 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. 

The corresponding sides of similar triangles are in proportion, so we can set up and solve the proportion statement for :

, so

For the sake of simplicuty, we will let 

Since  is the midpoint of , .

Also, .

The proportion statement becomes

Solve for  using cross-products:

By the quadratic equation, setting :

There are two possibilities:

or

 is divided into segments of length 2.9 and 17.1. The lesser is the length of , so the correct choice is 2.9.

Begin by distributing the three on the right side of the equation: 

Next combine your like terms by subtracting  from both sides to give you 

Next, subtract 9 from both sides to give you . To solve for , now take the square root of both sides. This gives you the answer, 

Given a quadratic equation equal to zero you can factor the equation and set each factor equal to zero. To factor you have to find two numbers that multiply to make 9 and add to make 6. The number is 3. So the factored form of the problem is (x+3)(x+3)=0. This statement is true only when x+3=0. Solving for x gives x=-3. Since this problem is multiple choice, you could also plug the given answers into the equation to see which one works. 

64x² + 24x - 10 = 0

Factor the equation:

(8x + 5)(8x – 2) = 0

Set each side equal to zero

(8x + 5) = 0

x = -5/8

(8x – 2) = 0

x = 2/8 = 1/4

The roots of an equation are the points at which the function equals zero. We can set each function equal to zero and determine which functions have one root, and which does not. 

Another piece of information will help. If a quadratic function has one root, then it must be a perfect square. This is because a quadratic function that is a perfect square can be written in the form (x – a)². If we set (x – a)² = 0 in order to find the root, we see that a is the only value that solves the equation, and thus a is the only root. Additionally, a quadratic equation is a perfect square if it can be written in the form a²x² + 2abx + b² = (ax + b)². 

Let's examine the choice f(x) = 4x² – 4x+1. To find the roots, we set f(x) = 0.

4x² – 4x+1 = 0

We notice that 4x² - 4x + 1 is a perfect square, since we could write it as (2x – 1)². Thus, this equation has only one root, and it can't be the answer.

If we look at f(x) = x² –2x + 1, we see that x² – 2x + 1 is also a perfect square, because it could be written as (x – 1)². This function also has a single root.

Next, we examine f(x) = (1/4)x² + x + 1. Let us set f(x) = (1/4)x² + x + 1 = 0.

(1/4)x² + x + 1 = 0

We can multiply both sides by four to get rid of the fraction.

x² + 4x + 4 = 0

(x + 2)² = 0

This function is also a perfect square and has a single root.

Now consider the choice f(x) = (–1/9)x² + 6x – 81.

f(x) = (–1/9)x² + 6x – 81 = 0

Multiply both sides by –9.

x² – 54x + 729 = 0

(x – 27)² = 0.

Finally, let's look at f(x) = 9x² – 6x + 4. This CANNOT be written as a perfect square, because it is not in the form a²x² + 2abx + b² = (ax + b)². It might be tempting to think that 9x² - 6x + 4 = (3x - 2)², but it does NOT, because (3x – 2)² = 9x² – 12x + 4. Therefore, because 9x² – 6x + 4 is not a perfect square, it doesn't have exactly one root.