### All PSAT Math Resources

## Example Questions

### Example Question #1 : Quadratic Equations

Solve: x^{2}+6x+9=0

**Possible Answers:**

12

6

9

3

-3

**Correct answer:**

-3

Given a quadratic equation equal to zero you can factor the equation and set each factor equal to zero. To factor you have to find two numbers that multiply to make 9 and add to make 6. The number is 3. So the factored form of the problem is (x+3)(x+3)=0. This statement is true only when x+3=0. Solving for x gives x=-3. Since this problem is multiple choice, you could also plug the given answers into the equation to see which one works.

### Example Question #2 : Quadratic Equations

64x^{2} + 24x - 10 = 0

Solve for x

**Possible Answers:**

**Correct answer:**1/4 and -5/8

64x^{2} + 24x - 10 = 0

Factor the equation:

(8x + 5)(8x – 2) = 0

Set each side equal to zero

(8x + 5) = 0

x = -5/8

(8x – 2) = 0

x = 2/8 = 1/4

### Example Question #3 : Quadratic Equations

All of the following functions have a exactly one root EXCEPT:

**Possible Answers:**

f(x) = (1/4)x^{2} + x + 1

f(x) = (–1/9)x^{2} + 6x – 81

f(x) = 4x^{2 }– 4x+1

f(x) = x^{2} – 2x + 1

f(x) = 9x^{2} – 6x + 4

**Correct answer:**

f(x) = 9x^{2} – 6x + 4

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 a^{2}x^{2} + 2abx + b^{2} = (ax + b)^{2}.

Let's examine the choice f(x) = 4x^{2 }– 4x+1. To find the roots, we set f(x) = 0.

4x^{2 }– 4x+1 = 0

We notice that 4x^{2 }- 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) = x^{2} –2x + 1, we see that x^{2} – 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)x^{2} + x + 1. Let us set f(x) = (1/4)x^{2} + x + 1 = 0.

(1/4)x^{2} + x + 1 = 0

We can multiply both sides by four to get rid of the fraction.

x^{2 }+ 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)x^{2} + 6x – 81.

f(x) = (–1/9)x^{2} + 6x – 81 = 0

Multiply both sides by –9.

x^{2 }– 54x + 729 = 0

(x – 27)^{2} = 0.

Finally, let's look at f(x) = 9x^{2} – 6x + 4. This CANNOT be written as a perfect square, because it is not in the form a^{2}x^{2} + 2abx + b^{2} = (ax + b)^{2}. It might be tempting to think that 9x^{2} - 6x + 4 = (3x - 2)^{2}, but it does NOT, because (3x – 2)^{2} = 9x^{2} – 12x + 4. Therefore, because 9x^{2} – 6x + 4 is not a perfect square, it doesn't have exactly one root. ^{}

### Example Question #4 : Quadratic Equations

The difference between a number and its square is 72. What is the number?

**Possible Answers:**

30

9

18

14

19

**Correct answer:**

9

x^{2} – x = 72. Solve for x using the quadratic formula and x = 9 and –8. Only 9 satisfies the restrictions.

### Example Question #1 : Quadratic Equations

Given and , find the value of .

**Possible Answers:**

**Correct answer:**

We can factor the quadratic equation into .

Then we can see that .

Therefore, becomes and .

### Example Question #1 : Quadratic Equations

Which of the following is a root of the function ?

**Possible Answers:**

**Correct answer:**

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 2*x* from the first two terms.

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

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

Subtract 1 from both sides.

2*x* = –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 .

### Example Question #7 : Quadratic Equations

36x^{2} -12x - 15 = 0

Solve for x

**Possible Answers:**

-1/2 and 5/6

1/2 and 1/3

1/2 and -1/3

-1/2 and -5/6

1/2 and 5/6

**Correct answer:**

-1/2 and 5/6

36x^{2} - 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

### Example Question #31 : Quadratic Equations

**Possible Answers:**

**Correct answer:**

### Example Question #32 : Quadratic Equations

Factor the following:

**Possible Answers:**

**Correct answer:**

Since you have an addition for the last element, you know that you will have either two subtractions or two additions in your factored groups. Since the second term is positive, your factored groups will contain two additions:

Now, all you have to do are go through the factors of 30 and find the pair that adds up to 11:

and : No!

and : No!

and : No! (But closer!)

and : Yes!

Therefore, your answer will be:

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