Irrational Numbers - SAT Subject Test in Math I

The first step is to distribute which gives us:

which is in standard form.

Combine like terms. Treat as if it were any other variable.

Substitute to eliminate .

Simplify.

A root of an integer is one of two things, an integer or an irrational number. By testing all five on a calculator, only comes up an exact integer - 5. This is the correct choice.

Multiply the numerator and denominator by the numerator's complex conjugate.

Reduce/simplify.

Use the FOIL technique:

We can set in the cube of a binomial pattern:

Multiply the numerator and the denominator by the conjugate of the denominator, which is . Then take advantage of the distributive properties and the difference of squares pattern:

Remember that

So the powers of are cyclic.This means that when we try to figure out the value of an exponent of , we can ignore all the powers that are multiples of because they end up multiplying the end result by , and therefore do nothing.

This means that

Now, remembering the relationships of the exponents of , we can simplify this to:

Because the elements on the left and right have to correspond (no mixing and matching!), we get the relationships:

No matter how you solve it, you get the values , .

To solve for , we must first solve the equation with the complex number for and . We therefore need to match up the real portion of the compex number with the real portions of the expression, and the imaginary portion of the complex number with the imaginary portion of the expression. We therefore obtain:

and

We can use substitution by noticing the first equation can be rewritten as and substituting it into the second equation. We can therefore solve for :

With this value, we can solve for :

Since we now have and , we can solve for :

Our final answer is therefore

To divide by a complex number, we must transform the expression by multiplying it by the complex conjugate of the denominator over itself. In the problem, is our denominator, so we will multiply the expression by to obtain:

.

We can then combine like terms and rewrite all terms as . Therefore, the expression becomes:

Our final answer is therefore

Multiply these complex numbers out in the typical way:

and recall that by definition. Then, grouping like terms we get

which is our final answer.

When dealing with imaginary numbers, we multiply by foiling as we do with binomials. When we do this we get the expression below:

Since we know that we get which gives us .

To add complex numbers, find the sum of the real terms, then find the sum of the imaginary terms.

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

Remember that , so .

Substitute in for

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

Remember that , so .

Substitute in for

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

Remember that , so .

Substitute in for

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

Remember that , so .

Substitute in for .

To get rid of the fraction, multiply the numerator and denominator by the conjugate of the denominator.

Now, multiply and simplify.

Remember that