All PSAT Math Resources
Example Questions
Example Question #3 : How To Find A Solution Set
Which of the following is true of the solution set of the equation ?
The solution set comprises one irrational number.
The solution set comprises one rational number.
The solution set comprises two rational numbers.
The solution set comprises two imaginary numbers.
The solution set comprises two irrational numbers.
The solution set comprises two irrational numbers.
First, since the equation is quadratic, put it in standard form
as follows:
To determine the nature of the solution set, evaluate discriminant for :
The discriminant is positive, but not a perfect square, so the solution set comprises two irrational numbers.
Example Question #1 : How To Find A Solution Set
Which of the following is true of the solution set of the equation ?
The solution set comprises two rational numbers and two imaginary numbers.
The solution set comprises two irrational numbers.
The solution set comprises four irrational numbers.
The solution set comprises two irrational numbers and two imaginary numbers.
The solution set comprises four imaginary numbers.
The solution set comprises two irrational numbers and two imaginary numbers.
Factor the polynomial as the difference of squares:
We set each binomial equal to 0 and apply the Square Root Property:
This yields two imaginary solutions.
This yields two irrational solutions.
The correct response is that the solution set comprises two irrational numbers and two imaginary numbers.
Example Question #4 : How To Find A Solution Set
Which of the following is true of the solution set of the equation ?
The solution set comprises one irrational number and two imaginary numbers.
The solution set comprises three irrational numbers.
The solution set comprises three imaginary numbers.
The solution set comprises one rational number and two irrational numbers.
The solution set comprises one rational number and two imaginary numbers.
The solution set comprises one rational number and two imaginary numbers.
The cubic binomial in the equation can be factored to yield an equivalent equation as follows::
One of these factors must be equal to 0.
If then , so one solution is rational.
If , we can find out about the nature of the remaining solutions using discriminant , setting :
The discriminant is negative, so the two solutions of the equation are imaginary. These are also the two remaining solutions of .
The correct response is that the equation has one rational solution and two imaginary solutions.
Example Question #21 : Equations / Solution Sets
J = a set of positive integer factors of 16
K = a set of positive integer factors of 24
L = a set of positive integer factors of 30
, , and represent three sets of numbers. What is the set of numbers that belongs in but not in or ?
This problem is asking for the factors of 24 that are NOT also factors of 16 OR 30. Thus, the quickest way to solve this problem is to just list all the factors of 16, 24, and 30, and solve. Those factors are:
16: 1, 2, 4, 8, 16
24: 1, 2, 3, 4, 6, 8, 12, 24
30: 1, 2, 3, 5, 6, 10, 15, 30
As you can see you can knock out 1 (16 & 30), 2 (16 & 30), 3 (30), 4 (16), 6 (30) and 8 (16) as factors of 24 that are in common with 16 & 30. This leaves only 12 and 24, which is the solution.
Example Question #1 : How To Find A Solution Set
If , what is the solution set for ?
To find the solution set, you must solve the equation; in this case, solving the equation means isolating on one side of the equation, and the numbers on the other side of the equation.
That is done like this:
K = -9 or 9 because either number is the square root of 81. To see that that's true, square both numbers. and .
This is very important to remember: whenever you're isolating a variable by taking the square root of a squared number, the answer can be a positive OR negative value, as long as they share an absolute value!
Example Question #1 : How To Find A Solution Set
Which of the following is true of the solution set of the equation ?
The solution set comprises two imaginary numbers and two irrational numbers.
The solution set comprises two imaginary numbers.
The solution set comprises two irrational numbers.
The solution set comprises four imaginary numbers.
The solution set comprises two imaginary numbers and two rational numbers.
The solution set comprises two imaginary numbers.
The perfect square trinomial in the equation can be factored to yield an equivalent equation as follows:
Therefore, there are exactly two solutions to the equation, both imaginary.
Example Question #11 : How To Find A Solution Set
Which of the following is true of the solution set of the equation ?
The solution set comprises two irrational numbers.
The solution set comprises two rational numbers.
The solution set comprises one irrational number.
The solution set comprises two imaginary numbers.
The solution set comprises one rational number.
The solution set comprises two imaginary numbers.
First, since the equation is quadratic, put it in standard form
as follows:
To determine the nature of the solution set, evaluate discriminant for :
The discriminant is negative, so the solution set comprises two imaginary numbers.
Example Question #161 : Expressions & Equations
Solve the system for and .
The most simple method for solving systems of equations is to transform one of the equations so it allows for the canceling out of a variable. In this case, we can multiply by to get .
Then, we can add to this equation to yield , so .
We can plug that value into either of the original equations; for example, .
So, as well.
Example Question #31 : Equations / Solution Sets
What is the solution to the following system of equations:
By solving one equation for , and replacing in the other equation with that expression, you generate an equation of only 1 variable which can be readily solved.
Example Question #113 : Algebra
Jeff, the barista at Moonbucks Coffee, is having a problem. He needs to make fifty pounds of Premium Blend coffee by mixing together some Kona beans, which cost $24 per pound, with some Ethiopian Delight beans, which cost $10 per pound. The Premium Blend coffee will cost $14.20 per pound. Also, the coffee will sell for the same price mixed as it would separately.
How many pounds of Kona beans will be in the mixture?
The number of pounds of coffee beans totals 50, so one of the equations would be
.
The total price of the Kona beans, is its unit price, $24 per pound, multiplied by its quantity, pounds. This is dollars. Similarly, the total price of the Ethiopian delight beans is dollars, and the price of the mixture is dollars. Add the prices of the Kona and Ethiopian Delight beans to get the price of the mixture:
We are trying to solve for in the system
Multiply the second equation by , then add to the first:
The mixture includes 15 pounds of Kona beans.
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