Algebra II - Math

Divide both sides by 3.

Consider both the negative and positive values for the absolute value term.

Subtract 2 from both sides to solve both scenarios for .

To solve absolute value equations, we must understand that the absoute value function makes a value positive. So when we are solving these problems, we must consider two scenarios, one where the value is positive and one where the value is negative.

and

This gives us:

and

However, this question has an outside of the absolute value expression, in this case . Thus, any negative value of will make the right side of the equation equal to a negative number, which cannot be true for an absolute value expression. Thus, is an extraneous solution, as cannot equal a negative number.

Our final solution is then

Notice that the equation has an term both inside and outside the absolute value expression.

Since the absolute value expression will always produce a positive number and the right side of the equation is negative, a negative number must be added to the result of the absolute value expression to satisfy the equation. Therefore the term outside of the absolute value expression (in this case ) must be negative (meaning must be negative).

Since will be a negative number, the expression within the absolute value will also be negative (before the absolute value is taken). It is thus possible to convert the original equation into an equation that treats the absolute value as a parenthetical expression that will be multiplied by , since any negative value becomes its opposite when taking the absolute value.

Simplifying and solving this equation for gives the answer:

The absolute value measures the distance from zero to the given point.

In this case, since , or , as both values are twelve units away from zero.

The absolute value can never be negative, so the equation is ONLY valid at zero.

The equation to solve becomes .

Simplifying this equation we notice that the 3's, 2's, and 1's cancel so

Alternative Solution

This expression can be simplified because all terms in the expression for 8! are also found in the expression for 10!. By writing the expression below, we are able to cancel 8!.

This is a factorial question. The formula for factorials is .

The cancels out all of except for the parts higher than 4, this leaves a 6 and a 5 left to multilpy

Both the numerator and denominator are factorials. If you expanded both, everything would cancel out except for in the numerator. Multiply those together to get 720.

Thus, since the remaining terms cancel out. 56 is the simplified result.

Remember a number followed by a ! is a factorial. A factorial is the product of the given number and all of the numbers smaller than it down to zero. For example, .

Rather than do all of the math involved for , notice that is the same as

From here, the 's cancel out, leaving us with .

To solve this equation, we have to factor our radicals. We do this by finding numbers that multiply to give us the number within the radical.

Add them together:

4 is a perfect square, so we can find the root:

Since both have the same radical, we can combine them:

Use the multiplication property of radicals to split the fourth roots as follows:

Simplify the new roots:

Begin by factoring the integer:

Now, simplify the exponents:

Begin by converting the radical into exponent form:

Now, multiply the exponents:

Begin by converting the radical into exponent form:

Now, combine the bases:

Simplify the integer:

Now, simplify the exponents:

Convert back into radical form and simplify:

Begin by using the FOIL method (First Outer Inner Last) to expand the expression.

Now, combine like terms:

Begin by multiplying the numerator and denominator by the complement of the denominator:

Use the FOIL method to multiply the radicals. F (first) O (outer) I (inner) L (last)

Now, combine like terms:

Simplify: