PSAT Math : Equations / Inequalities

Study concepts, example questions & explanations for PSAT Math

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Example Questions

Example Question #571 : Algebra

Two consecutive positive multiples of three have a product of 504.  What is the sum of the two numbers?

Possible Answers:

Correct answer:

Explanation:

Let  = the first positive number and  = the second positive number.

So the equation to solve is

We multiply out the equation and set it equal to zero before factoring.

x^{2} + 3x - 504 = 0 thus the two numbers are 21 and 24 for a sum of 45.

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

Two consecutive positive numbers have a product of 420.  What is the sum of the two numbers?

Possible Answers:

Correct answer:

Explanation:

Let  = first positive number and  = second positive number

So the equation to solve becomes

Using the distributive property, multiply out the equation and then set it equal to 0.  Next factor to solve the quadratic.

Example Question #31 : How To Find The Solution To A Quadratic Equation

A rectangle has a perimeter of 50\ m and an area of 150\ m^{2}   What is the difference between the length and width?

Possible Answers:

15\ m

5\ m

20\ m

10\ m

25\ m

Correct answer:

5\ m

Explanation:

For a rectangle, P=2w+2l and A=lw where w = width and l = length.

So we get two equations with two unknowns:

50=2w+2l

25=w+l 

l=25-w

150=lw

Making a substitution we get

150=(25-w)w

w^{2} -25w + 150 = 0

Solving the quadratic equation we get w=10\ m or 15\ m.

l=15\ m\ or\ 10\ m

The difference is 5\ m.

Example Question #181 : Algebra

If then which of the following is a possible value for ?

Possible Answers:

Correct answer:

Explanation:

  

 

Since , .

Thus

Of these two, only 4 is a possible answer.

Example Question #11 : How To Find The Solution To A Quadratic Equation

Find all real solutions to the equation.

Possible Answers:

Correct answer:

Explanation:

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 #31 : How To Find The Solution To A Quadratic Equation

How many real solutions are there for the following equation?

Possible Answers:

Correct answer:

Explanation:

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:

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 #31 : Quadratic Equations

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 #161 : Equations / Inequalities

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 #251 : Algebra

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, 

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