Absolute Value - SAT Math

When solving for x in the presence of absolute value, there are always two answers.

To eliminate the absolute value, the equation must be re-written two ways:

and

and

and

and

and

Substitute – 4 in for x. Remember that when a negative number is raised to the third power, it is negative. - = – 64. – 64 – 36 = – 100. Since you are asked to take the absolute value of – 100 the final value of f(-4) = 100. The absolute value of any number is positive.

We need to solve the two equations |2a – 3| = 5 and |3 – 4b| = 11, in order to determine the possible values of a and b. When solving equations involving absolute values, we must remember to consider both the positive and negative cases. For example, if |x| = 4, then x can be either 4 or –4.

Let's look at |2a – 3| = 5. The two equations we need to solve are 2a – 3 = 5 and 2a – 3 = –5.

2a – 3 = 5 or 2a – 3 = –5

Add 3 to both sides.

2a = 8 or 2a = –2

Divide by 2.

a = 4 or a = –1

Therefore, the two possible values for a are 4 and –1. However, the problem states that both a and b are negative. Thus, a must equal –1.

Let's now find the values of b.

3 – 4b = 11 or 3 – 4b = –11

Subtract 3 from both sides.

–4b = 8 or –4b = –14

Divide by –4.

b = –2 or b = 7/2

Since b must also be negative, b must equal –2.

We have determined that a is –1 and b is –2. The original question asks us to find |b – a|.

|b – a| = |–2 – (–1)| = | –2 + 1 | = |–1| = 1.

The answer is 1.

When we take the absolute value of a function, any negative values get changed into positive values. Essentially, |f(x)| will take all of the negative values of f(x) and reflect them across the x-axis. However, any values of f(x) that are positive or equal to zero will not be changed, because the absolute value of a positive number (or zero) is still the same number.

If we can show that f(x) has negative values, then |f(x)| will be different from f(x) at some points, because its negative values will be changed to positive values. In other words, our answer will consist of the function that never has negative values.

Let's look at f(x) = 2_x_ + 3. Obviously, this equation of a line will have negative values. For example, where x = –4, f(–4) = 2(–4) + 3 = –5, which is negative. Thus, f(x) has negative values, and if we were to graph |f(x)|, the result would be different from f(x). Therefore, f(x) = 2_x_ + 3 isn't the correct answer.

Next, let's look at f(x) = _x_2 – 9. If we let x = 1, then f(1) = 1 – 9 = –8, which is negative. Thus |f(x)| will not be the same as f(x), and we can eliminate this choice as well.

Now, let's examine f(x) = x_2 – 2_x. We know that _x_2 by itself can never be negative. However, if x_2 is really small, then adding –2_x could make it negative. Therefore, let's evaluate f(x) when x is a fractional value such as 1/2. f(1/2) = 1/4 – 1 = –3/4, which is negative. Thus, there are some values on f(x) that are negative, so we can eliminate this function.

Next, let's examine f(x) = _x_4 + x. In general, any number taken to an even-numbered power must always be non-negative. Therefore, _x_4 cannot be negative, because if we multiplied a negative number by itself four times, the result would be positive. However, the x term could be negative. If we let x be a small negative fraction, then _x_4 would be close to zero, and we would be left with x, which is negative. For example, let's find f(x) when x = –1/2. f(–1/2) = (–1/2)4 + (–1/2) = (1/16) – (1/2) = –7/16, which is negative. Thus, |f(x)| is not always the same as f(x).

By process of elimination, the answer is f(x) = _x_4 + (1 – x)2 . This makes sense because _x_4 can't be negative, and because (1 – x)2 can't be negative. No matter what we subtract from one, when we square the final result, we can't get a negative number. And if we add _x_4 and (1 – x)2, the result will also be non-negative, because adding two non-negative numbers always produces a non-negative result. Therefore, f(x) = _x_4 + (1 – x)2 will not have any negative values, and |f(x)| will be the same as f(x) for all values of x.

The answer is f(x) = _x_4 + (1 – x)2 .

To solve, we replace each variable with the given value.

Simplify. Remember that terms inside of the absolute value are always positive.

Substitute 0.6 for :

Substitute .

is the absolute value of , which in turn is the sum of a number and seven and a number. Therefore, can be written as "the absolute value of the sum of a number and seven". Since it is equal to , it is three less than the number, so the equation that corresponds to the sentence is

"The absolute value of the sum of a number and seven is three less than the number."

Simplify the following:

Begin with basic subtraction:

Next, remember what we do with absolute value signs; we change negative values to positive values, and positive values remain positive.

So our answer is positive 51

, or, equivalently,

, or, equivalently,

In order the find the answer, you must first solve what is inside the absolute value signs.

Following order of operations, you must first multiply which equals .

Then you must subtract from as shown below:

Now, you must take the absolute value of which is positive , the correct answer.

Since this is an absolute value equation, we must set the left hand side equal to both the positive and negative versions of the right side. Therefore,

Solving the first equation, we see that

Solving the second, we see that

When we substitute each value back into the original equation , we see that they both check.

To solve this equation, we want to set equal to both and and solve for .

Therefore:

and

We can check both of these answers by plugging them back into the inequality to see if the statement is true.

and

Both answers check, so our final answer is