### All SAT Math Resources

## Example Questions

### Example Question #19 : Quadratic Equations

Find all real solutions to the equation.

**Possible Answers:**

**Correct 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 .

### Example Question #20 : Quadratic Equations

How many real solutions are there for the following equation?

**Possible Answers:**

**Correct answer:**

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.

### Example Question #31 : Quadratic Equations

The formula to solve a quadratic expression is:

All of the following equations have real solutions EXCEPT:

**Possible Answers:**

**Correct answer:**

We can use the quadratic formula to find the solutions to quadratic equations in the form ax^{2 }+ 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, b^{2 }– 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 b^{2 }–^{ }4ac. If the value of b^{2 }– 4ac is negative, then this equation will not have real solutions.

Let's look at the equation 2x^{2} – 4x + 5 = 0 and determine the value of b^{2 }– 4ac.

b^{2 }– 4ac = (–4)^{2} – 4(2)(5) = 16 – 40 = –24 < 0

Because the value of b^{2 }– 4ac is less than zero, this equation will not have real solutions. Our answer will be 2x^{2} – 4x + 5 = 0.

If we inspect all of the other answer choices, we will find positive values for b^{2 }– 4ac, and thus these other equations will have real solutions.

### Example Question #32 : Quadratic Equations

Let , and let . What is the sum of the possible values of such that .

**Possible Answers:**

**Correct answer:**

We are told that f(x) = x^{2} - 4x + 2, and g(x) = 6 - x. Let's find expressions for f(k) and g(k).

f(k) = k^{2} – 4k + 2

g(k) = 6 – k

Now, we can set these expressions equal.

f(k) = g(k)

k^{2} – 4k +2 = 6 – k

Add k to both sides.

k^{2} – 3k + 2 = 6

Then subtract 6 from both sides.

k^{2} – 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.

### Example Question #32 : Quadratic Equations

Solve for x: x^{2} + 4x = 5

**Possible Answers:**

-1

None of the other answers

-1 or 5

-5

-5 or 1

**Correct answer:**

-5 or 1

Solve by factoring. First get everything into the form Ax^{2} + Bx + C = 0:^{}

x^{2} + 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

### Example Question #32 : Quadratic Equations

Solve for x: (x^{2} – x) / (x – 1) = 1

**Possible Answers:**

x = -1

x = -2

x = 2

No solution

x = 1

**Correct answer:**

No solution

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

x^{2} – x = x – 1

Solve as a quadratic equation: x^{2} – 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.

### Example Question #33 : Quadratic Equations

A farmer has 44 feet of fence, and wants to fence in his sheep. He wants to build a rectangular pen with an area of 120 square feet. Which of the following is a possible dimension for a side of the fence?

**Possible Answers:**

**Correct answer:**

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.

### Example Question #61 : New Sat Math Calculator

Find the roots of the following equation:

**Possible Answers:**

**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:

### Example Question #34 : Quadratic Equations

Give the solution set of the equation

**Possible Answers:**

The equation has no solution.

**Correct answer:**

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 .

### Example Question #37 : Quadratic Equations

Consider the equation

.

Which of the following statements correctly describes its solution set?

**Possible Answers:**

Exactly two solutions, both of which are imaginary.

Exactly two solutions, both of which are irrational.

Exactly one solution, which is irrational.

Exactly one solution, which is rational.

Exactly two solutions, both of which are rational.

**Correct answer:**

Exactly two solutions, both of which are 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 positive, so there are two real solutions.

529 is a perfect square:

.

Therefore, the two real solutions are also rational.

Certified Tutor