Let  = the first positive number and  = the second positive number.

So the equation to solve is

We multiply out the equation and set it equal to zero before factoring.

thus the two numbers are 21 and 24 for a sum of 45.

Let  = first positive number and  = second positive number

So the equation to solve becomes

Using the distributive property, multiply out the equation and then set it equal to 0.  Next factor to solve the quadratic.

For a rectangle,  and  where  = width and  = length.

So we get two equations with two unknowns:

Making a substitution we get

Solving the quadratic equation we get  or .

The difference is .

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. 

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 .

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,