There are two values where x² = x, namely x = 0 and x = 1. All values between 0 and 1 get smaller after squaring. All values greater than 1 get larger after squaring.

In order to find the value of f(g(k)), we will first need to find k. We are told that f(k) = 31, so we can write an expression for f(k) and solve for k.

f(x) = 2x² – 4x + 1

f(k) = 2k² – 4k + 1 = 31

Subtract 31 from both sides.

2k² – 4k – 30 = 0

Divide both sides by 2.

k² – 2k – 15 = 0

Now, we can factor this by thinking of two numbers that multiply to give –15 and add to give –2. These two numbers are –5 and 3.

k² –2k – 15 = (k – 5)(k + 3) = 0

We can set each factor equal to 0 to find the values for k.

k – 5 = 0

Add 5 to both sides.

k = 5

Now we set k + 3 = 0.

Subtract 3 from both sides.

k = –3

This means that k could be either 5 or –3. However, we are told that k is a negative number, which means k = –3.

Finally, we can evaluate the expression f(g(–3)). First we need to find g(–3).

g(x) = (x² + 16)^(1/2)

g(–3) = ((–3)² + 16)^(1/2)

= (9 + 16)^(1/2)

= 25^(1/2)

Raising something to the one-half power is the same as taking the square root.

25^(1/2) = 5

Now that we know g(–3) = 5, we must find f(5).

f(5) = 2(5)² – 4(5) + 1

= 2(25) – 20 + 1 = 31

The answer is 31.

The equation in the problem is quadratic, so we can use the quadratic formula to solve it. If an equation is in the form ax² + bx + c = 0, where a, b, and c are constants, then the quadratic formula, given below, gives us the solutions of x.

In this particular problem, a = 2, b = –6, and c = 3.

The value under the square-root, b² – 4ac, is called the discriminant, and it gives us important information about the nature of the solutions of a quadratic equation.

If the discriminant is less than zero, then the roots are not real, because we would be forced to take the square root of a negative number, which yields an imaginary result. The discriminant of the equation we are given is (–6)² – 4(2)(3) = 36 – 24 = 12 > 0. Because the discriminant is not negative, the solutions to the equation will be real. Thus, option I is correct.

The discriminant can also tell us whether the solutions of an equation are rational or not. If we take the square root of the discriminant and get a rational number, then the solutions of the equation must be rational. In this problem, we would need to take the square root of 12. However, 12 is not a perfect square, so taking its square root would produce an irrational number. Therefore, the solutions to the equation in the problem cannot be rational. This means that choice II is incorrect.

Lastly, the discriminant tells us if the roots to an equation are distinct (different from one another). If the discriminant is equal to zero, then the solutions of x become (–b + 0)/2a and (–b – 0)/2a, because the square root of zero is 0. Notice that (–b + 0)/2a is the same as (–b – 0)/2a. Thus, if the discriminant is zero, then the roots of the equation are the same, i.e. indistinct. In this particular problem, the discriminant = 12, which doesn't equal zero. This means that the two roots will be different, i.e. distinct. Therefore, choice III applies.

The answer is choices I and III only.

First let's see if there is a common term. 

3x² + 15x – 18 = 0

We can pull out a 3: 3(x² + 5x – 6) = 0

Divide both sides by 3: x² + 5x – 6 = 0

We need two numbers that sum to 5 and multiply to –6. 6 and –1 work.

(x + 6)(x – 1) = 0

x = –6 or x = 1

Factor the expression and set each factor equal to 0:

Let  = the first number and  = the second number.

So the equation to solve becomes .  This quadratic equation needs to be multiplied out and set equal to zero before factoring.  Then set each factor equal to zero and solve.  Only positive numbers are correct, so the answer is .

Let  = first positive number and  = second positive number. 

The equation to solve becomes

We multiply out this quadratic equation and set it equal to 0, then factor.

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

The equation to solve becomes

We solve this quadratic equation by multiplying it out and setting it equal to 0.  The next step is to factor.

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.