## Example Questions

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### Example Question #2 : Quadratic Equation

If f(x) = -x2 + 6x - 5, then which could be the value of a if f(a) = f(1.5)?

2.5
1
3.5
4.5
4
Explanation:

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.

### Example Question #1 : How To Find The Solution To A Quadratic Equation

Solve for x:  2(x + 1)2 – 5 = 27

3 or –5

3 or 4

–3 or 2

–2 or 4

–2 or 5

3 or –5

Explanation:

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.

### Example Question #2 : How To Find The Solution To A Quadratic Equation

Two consecutive positive multiples of three have a product of 54. What is the sum of the two numbers?

6

3

15

9

12

15

Explanation:

Define the variables to be x = first multiple of three and x + 3 = the next consecutive multiple of 3.

Knowing the product of these two numbers is 54 we get the equation x(x + 3) = 54. To solve this quadratic equation we need to multiply it out and set it to zero then factor it. So x2 + 3x – 54 = 0 becomes (x + 9)(x – 6) = 0.  Solving for x we get x = –9 or x = 6 and only the positive number is correct.  So the two numbers are 6 and 9 and their sum is 15.

### Example Question #2 : Quadratic Equation

Solve 3x2 + 10x = –3

x = –1/6 or –6

x = –2/3 or –2

x = –1/3 or –3

x = –4/3 or –1

x = –1/9 or –9

x = –1/3 or –3

Explanation:

First, the equations must be put in standard form: 3x2 + 10x + 3 = 0

Second, try to factor the quadratic; however, if that is not possible use the quadratic formula.

Third, check the answer by plugging the answers back into the original equation.

### Example Question #11 : Quadratic Equations

3x2 – 11x = –10

Which of the following is a valid value for x?

None of the other answers

5 / 3

-2

-5 / 3

3

5 / 3

Explanation:

Begin by getting our equation into the form Ax2 + BX + C = 0:

3x2 – 11x + 10 = 0

Now, if you factor the left, you can find the answer. Begin by considering the two groups.  They will have to begin respectively with 3 and 1 as coefficients for your x value.  Likewise, looking at the last element, you can tell that both will have to have a + or –, since the C coefficient is positive.  Finally, since the B coefficient is negative, we know that it will have to be –. We know therefore:

(3x – ?)(x – ?)

The potential factors of 10 are: 10, 1; 1, 10; 2, 5; 5, 2

5 and 2 work:

(3x – 5)(x – 2) = 0 because you can FOIL (3x – 5)(x – 2) back into 3x2 – 11x + 10.

Now, the trick remaining is to set each of the factors equal to 0 because if either group is 0, the whole equation will be 0:

3x – 5 = 0 → 3x = 5 → x = 5/3

x – 2 = 0 → x = 2

Therefore, x is either 5 / 3 or 2. The former is presented as an answer.

### Example Question #1 : How To Find The Solution To A Quadratic Equation

What is the sum of the values of x that satisfy the following equation:

16x – 10(4)x + 16 = 0.

2

3/2

1

4

5/2

2

Explanation:

The equation we are asked to solve is 16x – 10(4)x + 16 = 0.

Equations of this type can often be "transformed" into other equations, such as linear or quadratic equations, if we rewrite some of the terms.

First, we can notice that 16 = 42. Thus, we can write 16x as (42)x or as (4x)2.

Now, the equation is (4x)2 – 10(4)x + 16 = 0

Let's introduce the variable u, and set it equal 4x. The advantage of this is that it allows us to "transform" the original equation into a quadratic equation.

u2 – 10u + 16 = 0

This is an equation with which we are much more familiar. In order to solve it, we need to factor it and set each factor equal to zero. In order to factor it, we must think of two numbers that multiply to give us 16 and add to give us –10. These two numbers are –8 and –2. Thus, we can factor u2 – 10u + 16 = 0 as follows:

(u – 8)(u – 2) = 0

Next, we set each factor equal to 0.

u – 8 = 0

u = 8

u – 2 = 0

u = 2.

Thus, u must equal 2 or 8. However, we want to find x, not u. Since we defined u as equal to 4x, the equations become:

4x = 2 or 4x = 8

Let's solve 4x = 2 first. We can rewrite 4x as (22)x = 22x, so that the bases are the same.

22x = 2 = 21

2x = 1

x = 1/2

Finally, we will solve 4x = 8. Once again, let's write 4x as 22x. We can also write 8 as 23.

22x = 23

2x = 3

x = 3/2

The original question asks us to find the sum of the values of x that solve the equation. Because x can be 1/2 or 3/2, the sum of 1/2 and 3/2 is 2.

The answer is 2.

### Example Question #1 : How To Find The Solution To A Quadratic Equation

If x > 0, what values of x satisfy the inequality x2x?

All positive real numbers

No values of x satifisfy the inequality

All perfect squares

All positive integers

All real numbers greater than one

All real numbers greater than one

Explanation:

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.

### Example Question #131 : Equations / Inequalities

Let f(x) = 2x2 – 4x + 1 and g(x) = (x2 + 16)(1/2). If k is a negative number such that f(k) = 31, then what is the value of (f(g(k))?

5

-81

31

25

-35

31

Explanation:

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.

2k– 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)(+ 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.

### Example Question #1 : How To Find The Solution To A Quadratic Equation

I. real

II. rational

III. distinct

Which of the descriptions characterizes the solutions of the equation 2x2 – 6x + 3 = 0?

I only

I and II only

I and III only

II only

II and III only

I and III only

Explanation:

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, b– 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.

### Example Question #31 : Quadratic Equations

Solve for x.

3x2 + 15x – 18 = 0.

x = –2 or x = 3

x = 6 or x = –1

x = –6 or x = 1

x = 5 or x = 1

x = 2 or x = –3

x = –6 or x = 1

Explanation:

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

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