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. 

Solve by factoring.  First get everything into the form Ax² + Bx + C = 0:

x² + 4x - 5 = 0

Then factor: (x + 5) (x - 1) = 0

Solve each multiple separately for 0:

X + 5 = 0; x = -5

x - 1 = 0; x = 1

Therefore, x is either -5 or 1

Begin by multiplying both sides by (x – 1):

x² – x = x – 1

Solve as a quadratic equation: x² – 2x + 1 = 0

Factor the left: (x – 1)(x – 1) = 0

Therefore,  x = 1.

However, notice that in the original equation, a value of 1 for x would place a 0 in the denominator.  Therefore, there is no solution.

Set up two equations from the given information:

 and 

Substitute into the second equation:

Multiply through by .

Then divide by the coefficient of 2 to simplify your work:

Then since you have a quadratic setup, move the  term to the other side (via subtraction from both sides) to set everything equal to 0:

As you look for numbers that multiply to positive 120 and add to -22 so you can factor the quadratic, you might recognize that -12 and -10 fit the bill. This makes your factorization:

This makes the possible solutions 10 and 12. Since 12 does not appear in the choices,  is the only possible correct answer.

The equation that is given can be factor into:

The roots is the locations where this equation equals zero as seen below:

This occurs when the value in either parenthesis equals zero.

Solving for the first expression:

Solving for the second root:

Therefore the roots are:

The quadratic trinomial can be factored using the  method by looking for two integers whose sum is  and whose product is . Throught trial and error, we see that these integers are  and 8, so we continue:

One of the factors must be equal to 0, so either:

or 

The correct choice is .

Write the quadratic equation in standard form  by subtracting  from both sides:

The nature of the solution set of a quadratic equation in standard form can be determined by examining the discriminant . Setting :

The discriminant is positive, so there are two real solutions. 

529 is a perfect square: 

.

Therefore, the two real solutions are also rational.

Write the quadratic equation in standard form  by subtracting  from both sides:

The nature of the solution set of a quadratic equation in standard form can be determined by examining the discriminant . Setting :

The discriminant is negative, so there are two imaginary solutions. 

Write the quadratic equation in standard form  by subtracting 12 from both sides:

The nature of the solution set of a quadratic equation in standard form can be determined by examining the discriminant . Setting :

The discriminant is positive, so there are two real solutions. However, 145 is not a perfect square; the two solutions are therefore irrational.