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. 

x² – x = 72. Solve for x using the quadratic formula and x = 9 and –8. Only 9 satisfies the restrictions.

We can factor the quadratic equation into .

Then we can see that .

Therefore,  becomes  and .

The roots of a function are the x intercepts of the function. Whenever a function passes through a point on the x-axis, the value of the function is zero. In other words, to find the roots of a function, we must set the function equal to zero and solve for the possible values of x.

This is a quadratic trinomial. Let's see if we can factor it. (We could use the quadratic formula, but it's easier to factor when we can.)

Because the coefficient in front of the is not equal to 1, we need to multiply this coefficient by the constant, which is –4. When we mutiply 2 and –4, we get –8. We must now think of two numbers that will multiply to give us –8, but will add to give us –7 (the coefficient in front of the x term). Those two numbers which multiply to give –8 and add to give –7 are –8 and 1. We will now rewrite –7x as –8x + x.

We will then group the first two terms and the last two terms.

We will next factor out a 2x from the first two terms.

Thus, when factored, the original equation becomes (2x + 1)(x – 4) = 0.

We now set each factor equal to zero and solve for x.

Subtract 1 from both sides.

2x = –1

Divide both sides by 2.

Now, we set x – 4 equal to 0.

x – 4 = 0

Add 4 to both sides.

x = 4

The roots of f(x) occur where x = .

The answer is therefore  .

36x² - 12x - 15 = 0

Factor the equation:

(6x + 3)(6x - 5) = 0

Set each side equal to zero

6x + 3 = 0

x = -3/6 = -1/2

6x – 5 = 0

x = 5/6

To factor this, we need to find a pair of numbers that multiplies to 6 and sums to 5. The numbers 2 and 3 work. (2 * 3 = 6 and 2 + 3 = 5)

(x + 2)(x + 3) = 0

x = –2 or x = –3

To solve, begin by factoring the equation. We know that in our two factors, one will begin with  and one will begin with . Fiddling with different factors of  (we are looking for two numbers that, when multiplied by  and  separately, will add to ), we come to the following:

(If you are unsure, double check by expanding the equation to match the original)

Now, set the each factor equal to 0:

Do the same for the second factor:

Therefore, our two values of  are  and .

Alternatively, this problem can be solved by plugging each answer choice into the original equation and finding which set of numbers make the equation equate to 0.

To solve for , we need to factor the equation to identify its roots, or the values of  that make the equation equal .

We can factor out a  from each of the terms on the left side of the equation since they have that in common:

Then, we can factor the remaining quadratic portion of the equation.  and  have a difference of , suggesting that trying  and  might work:

At this point, we can identify the roots of the equation by setting  and  and solving each of these equations to yield  and , respectively.

We need to input 1.5 into our function, then we need to input "a" into our function and set these results equal.

f(a) = f(1.5)

f(a) = -(1.5)² +6(1.5) -5

f(a) = -2.25 + 9 - 5

f(a) = 1.75

-a² + 6a -5 = 1.75

Multiply both sides by 4, so that we can work with only whole numbers coefficients.

-4a² + 24a - 20 = 7

Subtract 7 from both sides.

-4a² + 24a - 27 = 0

Multiply both sides by negative one, just to make more positive coefficients, which are usually easier to work with.

4a² - 24a + 27 = 0

In order to factor this, we need to mutiply the outer coefficients, which gives us 4(27) = 108. We need to think of two numbers that multiply to give us 108, but add to give us -24. These two numbers are -6 and -18. Now we rewrite the equation as:

4a² - 6a -18a + 27 = 0

We can now group the first two terms and the last two terms, and then we can factor.

(4a² - 6a )+(-18a + 27) = 0

2a(2a-3) + -9(2a - 3) = 0

(2a-9)(2a-3) = 0

This means that 2a - 9 =0, or 2a - 3 = 0.

2a - 9 = 0

2a = 9

a = 9/2 = 4.5

2a - 3 = 0

a = 3/2 = 1.5

So a can be either 1.5 or 4.5.

The only answer choice available that could be a is 4.5.

Quadratic equations generally have two answers.  We add 5 to both sides and then divide by 2 to get the quadratic expression on one side of the equation: (x + 1)² = 16.   By taking the square root of both sides we get x + 1 = –4 or x + 1 = 4.  Then we subtract 1 from both sides to get x = –5 or x = 3.