Equations / Solution Sets - 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.

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.

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.

Since we know that stands for weeks, the answer has to have something to do with the weeks. This eliminates "the number of candles she has remaining for the month." Also, we can eliminate "the number of weeks that she has sold candles this month" because that would be our value for , not what we'd multiply by. The correct answer is, "the number of candles that she sells each week."

The correct answer is (x + 4)(x – 4)

We neen to factor x2 – 16 to solve. We know that each parenthesis will contain an x to make the x2. We know that the root of 16 is 4 and since it is negative and no value of x is present we can tell that one 4 must be positive and the other negative. If we work it from the multiple choice answers we will see that when multiplying it out we get x2 + 4x – 4x – 16. 4x – 4x cancels out and we are left with our answer.

First, let's factor x3 – y3 using the formula for difference of cubes.

x3 – y3 = (x – y)(x2 + xy + y2)

We are told that x2 + xy + y2 = 6. Thus, we can substitute 6 into the above equation and solve for x – y.

(x - y)(6) = 30.

Divide both sides by 6.

x – y = 5.

The original questions asks us to find x2 – 2xy + y2. Notice that if we factor x2 – 2xy + y2 using the formula for perfect squares, we obtain the following:

x2 – 2xy + y2 = (x – y)2.

Since we know that (x – y) = 5, (x – y)2 must equal 52, or 25.

Thus, x2 – 2xy + y2 = 25.

The answer is 25.

This can be solved by substitution and factoring.

x2 – y = 34 can be written as y = x2 – 34 and substituted into the other equation: x – y = 4 which leads to x – x2 + 34 = 4 which can be written as x2 – x – 30 = 0.

x2 – x – 30 = 0 can be factored to (x – 6)(x + 5) = 0 so x = 6 and –5 and because only 6 is a possible answer, it is the correct choice.

When a = 9, then x_2 + 2_ax + 81 = 0 becomes

x_2 + 18_x + 81 = 0.

This equation can be factored as (x + 9)2 = 0.

Therefore when a = 9, x = –9.

In general, if a function has a root at x = r, then (x – r) must be a factor of f(x). In this problem, we are told that f(x) has roots at –1, 0 and 2. This means that the following are all factors of f(x):

(x – (–1)) = x + 1

(x – 0) = x

and (x – 2).

This means that we must look for an equation for f(x) that has the factors (x + 1), x, and (x – 2).

We can immediately eliminate the function f(x) = _x_2 + x – 2, because we cannot factor an x out of this polynomial. For the same reason, we can eliminate f(x) = _x_2 – x – 2.

Let's look at the function f(x) = _x_3 – x_2 + 2_x. When we factor this, we are left with x(_x_2 – x + 2). We cannot factor this polynomial any further. Thus, x + 1 and x – 2 are not factors of this function, so it can't be the answer.

Next, let's examine f(x) = _x_4 + _x_3 – 2_x_2 .

We can factor out _x_2.

_x_2 (_x_2 + x – 2)

When we factor _x_2 + x – 2, we will get (x + 2)(x – 1). These factors are not the same as x – 2 and x + 1.

The only function with the right factors is f(x) = _x_3 – x_2 – 2_x.

When we factor out an x, we get (_x_2 – x – 2), which then factors into (x – 2)(x + 1). Thus, this function has all of the factors we need.

The answer is f(x) = _x_3 – x_2 – 2_x.

The numbers by which _x_6 – _y_6 is divisible will be all of its factors. In other words, we need to find all of the factors of _x_6 – _y_6 , which essentially means we must factor _x_6 – _y_6 as much as we can.

First, we will want to apply the difference of squares rule, which states that, in general, _a_2 – _b_2 = (a – b)(a + b). Notice that a and b are the square roots of the values of _a_2 and _b_2, because √_a_2 = a, and √_b_2 = b (assuming a and b are positive). In other words, we can apply the difference of squares formula to _x_6 – _y_6 if we simply find the square roots of _x_6 and _y_6.

Remember that taking the square root of a quantity is the same as raising it to the one-half power. Remember also that, in general, (ab)c = abc.

√_x_6 = (_x_6)(1/2) = x(6(1/2)) = _x_3

Similarly, √_y_6 = _y_3.

Let's now apply the difference of squares factoring rule.

_x_6 – _y_6 = (_x_3 – _y_3)(_x_3 + _y_3)

Because we can express _x_6 – _y_6 as the product of (_x_3 – _y_3) and (_x_3 + _y_3), both (_x_3 – _y_3) and (_x_3 + _y_3) are factors of _x_6 – _y_6 . Thus, we can eliminate _x_3 – _y_3 from the answer choices.

Let's continue to factor (_x_3 – _y_3)(_x_3 + _y_3). We must now apply the sum of cubes and differences of cubes formulas, which are given below:

In general, _a_3 + _b_3 = (a + b)(_a_2 – ab + _b_2). Also, _a_3 – _b_3 = (a – b)(_a_2 + ab + _b_2)

Thus, we have the following:

(_x_3 – _y_3)(_x_3 + _y_3) = (x – y)(_x_2 + xy + _y_2)(x + y)(_x_2 – xy + _y_2)

This means that x – y and x + y are both factors of _x_6 – _y_6 , so we can eliminate both of those answer choices.

We can rearrange the factorization (x – y)(_x_2 + xy + _y_2)(x + y)(_x_2 – xy + _y_2) as follows:

(x – y)(x + y)(_x_2 + xy + _y_2)(_x_2 – xy + _y_2)

Notice that (x – y)(x + y) is merely the factorization of difference of squares. Therefore, (x – y)(x + y) = _x_2 – _y_2.

(x – y)(x + y)(_x_2 + xy +_y_2)(_x_2 – xy + _y_2) = (_x_2 – _y_2)(_x_2 + xy +_y_2)(_x_2 – xy + _y_2)

This means that _x_2 – _y_2 is also a factor of _x_6 – _y_6.

By process of elimination, _x_2 + _y_2 is not necessarily a factor of _x_6 – _y_6 .

The answer is _x_2 + _y_2 .

First pull out any common terms: 4_x_3 – 16_x_ = 4_x_(_x_2 – 4)

_x_2 – 4 is a difference of squares, so we can also factor that further. The difference of squares formula is _a_2 – _b_2 = (a – b)(a + b). Here a = x and b = 2. So _x_2 – 4 = (x – 2)(x + 2).

Putting everything together, 4_x_3 – 16_x_ = 4_x_(x + 2)(x – 2).

This is a difference of squares. The difference of squares formula is a_2 – b_2 = (a + b)(a – b). In this problem, a = 6_x and b = 7_y.

So 36_x_2 – 49_y_2 = (6_x_ + 7_y_)(6_x_ – 7_y_).