PSAT Math : Quadratic Equations

Study concepts, example questions & explanations for PSAT Math

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Example Questions

Example Question #21 : How To Find The Solution To A Quadratic Equation

The formula to solve a quadratic expression is:

All of the following equations have real solutions EXCEPT:

Possible Answers:

Correct answer:

Explanation:

We can use the quadratic formula to find the solutions to quadratic equations in the form ax+ 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– 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 4ac. If the value of b– 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 b– 4ac.

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

Because the value of b– 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 b– 4ac, and thus these other equations will have real solutions.

Example Question #22 : How To Find The Solution To A Quadratic Equation

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

Possible Answers:

Correct answer:

Explanation:

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. 

Example Question #21 : Quadratic Equations

Stuff

Note: Figure NOT drawn to scale.

Refer to the above diagram, which shows Rectangle  with .

 is the midpoint of 

Evaluate  (to the nearest tenth, if applicable).

Possible Answers:

Insufficient information is given to answer the question.

Correct answer:

Explanation:

The corresponding sides of similar triangles are in proportion, so we can set up and solve the proportion statement for :

, so

 

For the sake of simplicuty, we will let 

Since  is the midpoint of .

Also, .

 

The proportion statement becomes

Solve for  using cross-products:

By the quadratic equation, setting :

There are two possibilities:

or

 is divided into segments of length 2.9 and 17.1. The lesser is the length of , so the correct choice is 2.9.

 

Example Question #22 : How To Find The Solution To A Quadratic Equation

Solve for

Possible Answers:

Correct answer:

Explanation:

Begin by distributing the three on the right side of the equation: 

Next combine your like terms by subtracting  from both sides to give you 

Next, subtract 9 from both sides to give you . To solve for , now take the square root of both sides. This gives you the answer, 

Example Question #1 : Quadratic Equations

Solve: x2+6x+9=0 

Possible Answers:

6

-3

3

9

12

Correct answer:

-3

Explanation:

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 #1 : Quadratic Equations

64x2 + 24x - 10 = 0

Solve for x

Possible Answers:
1/4 and 5/8
1/4 and -5/8
-1/4 and 3/4
1/4 and -3/4
-1/4 and -5/8
Correct answer: 1/4 and -5/8
Explanation:

64x2 + 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) = (–1/9)x2 + 6x – 81

f(x) = 4x– 4x+1

f(x) = (1/4)x2 + x + 1

f(x) = x2 – 2x + 1

f(x) = 9x2 – 6x + 4

Correct answer:

f(x) = 9x2 – 6x + 4

Explanation:

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) = 4x– 4x+1. To find the roots, we set f(x) = 0.

4x– 4x+1 = 0

We notice that 4x- 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.

x+ 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.

x– 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. 

Example Question #4 : Quadratic Equations

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

Possible Answers:

19

18

30

14

9

Correct answer:

9

Explanation:

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

Example Question #5 : Quadratic Equations

Given x^{2}-6x+9=0 and x\cdot k=1, find the value of k.

Possible Answers:

\frac{1}{3}

1

\frac{1}{2}

\frac{2}{3}

\frac{3}{2}

Correct answer:

\frac{1}{3}

Explanation:

We can factor the quadratic equation into (x-3)^{2}=0.

Then we can see that x=3.

Therefore, x\cdot k=1 becomes 3\cdot k=1 and k=\frac{1}{3}.

Example Question #2 : Quadratic Equations

Which of the following is a root of the function f(x)=2x^2-7x-4 ?

Possible Answers:

x = -4

x = -\frac {1}{2}

x = \frac{1}{4}

x = \frac{1}{2}

x = -2

Correct answer:

x = -\frac {1}{2}

Explanation:

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.

f(x)=2x^2-7x-4 = 0

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 x^2 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.

2x^2-7x-4=2x^2-8x+x-4=0

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

(2x^2-8x)+(x-4)=0

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

(2x^2-8x)+(x-4)=2x(x-4)+1(x-4)=(2x+1)(x-4)=0

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

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

2x + 1 = 0

Subtract 1 from both sides.

2x = –1

Divide both sides by 2.

x=-\frac{1}{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 = -\frac{1}{2},4.

The answer is therefore  x = -\frac {1}{2}.

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