How to find a solution set - PSAT Math

The original statement is equivalent to a statement that is identically true regardless of the value of ; therefore, so is the original statement itself. The solution set is the set of all real numbers.

To solve this equation, simply plug 12 in for in the equation.

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.

The question is equivalent to asking the following:

For what value of does the equation

have more than one solution?

The equation simplifies as follows:

If the absolute values of two expressions equal, then either the expressions themselves are equal or they are each other's opposite.

Taking the latter case:

Regardless of the value of , exactly one solution is yielded this way.

The question becomes as follows: for which value of does the other way yield a solution?

Set:

If this is a false statement, then this yields no solutions.

If this is a true statement, then this automatically yields the set of all real numbers as the solution set. We solve for :

As a result, the statement

has infinitely many solutions, and the other three statements have exactly one.

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.

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 and a perfect square, so the solution set comprises two rational 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.

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.

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.

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!

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.

The original statement is equivalent to a statement that is identically true regardless of the value of ; therefore, so is the original statement itself. The solution set is the set of all real numbers.

To solve this equation, simply plug 12 in for in the equation.

The question is equivalent to asking the following:

For what value of does the equation

have more than one solution?

The equation simplifies as follows:

If the absolute values of two expressions equal, then either the expressions themselves are equal or they are each other's opposite.

Taking the latter case:

Regardless of the value of , exactly one solution is yielded this way.

The question becomes as follows: for which value of does the other way yield a solution?

Set:

If this is a false statement, then this yields no solutions.

If this is a true statement, then this automatically yields the set of all real numbers as the solution set. We solve for :

As a result, the statement

has infinitely many solutions, and the other three statements have exactly one.

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.

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 and a perfect square, so the solution set comprises two rational 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.

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.

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.

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.