All PSAT Math Resources
Example Questions
Example Question #23 : Quadratic Equations
Two consecutive positive multiples of three have a product of 504. What is the sum of the two numbers?
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.
thus the two numbers are 21 and 24 for a sum of 45.
Example Question #21 : Quadratic Equations
Two consecutive positive numbers have a product of 420. What is the sum of the two numbers?
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 #24 : Quadratic Equations
A rectangle has a perimeter of and an area of What is the difference between the length and width?
For a rectangle, and where = width and = length.
So we get two equations with two unknowns:
Making a substitution we get
Solving the quadratic equation we get or .
The difference is .
Example Question #25 : Quadratic Equations
If then which of the following is a possible value for ?
Since , .
Thus
Of these two, only 4 is a possible answer.
Example Question #26 : Quadratic Equations
Find all real solutions to the equation.
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 #27 : Quadratic Equations
How many real solutions are there for the following equation?
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 #302 : Equations / Inequalities
The formula to solve a quadratic expression is:
All of the following equations have real solutions EXCEPT:
We can use the quadratic formula to find the solutions to quadratic equations in the form ax2 + 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, b2 – 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 b2 – 4ac. If the value of b2 – 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 b2 – 4ac.
b2 – 4ac = (–4)2 – 4(2)(5) = 16 – 40 = –24 < 0
Because the value of b2 – 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 b2 – 4ac, and thus these other equations will have real solutions.
Example Question #21 : Quadratic Equations
Let , and let . What is the sum of the possible values of such that .
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 : How To Find The Solution To A Quadratic Equation
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).
Insufficient information is given to answer the question.
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 #21 : How To Find The Solution To A Quadratic Equation
Solve for :
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,