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Solving Systems of Linear Equations in Three Variables
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Simplifying Sums and Differences of Square Roots
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Powers of a Monomial
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Solving Systems of Equations By Addition (Elimination)
The Product and Quotient Rules
Linear Systems of Equations with No Solution
Solving Quadratic Equations Using the Quadratic Formula
Solving Quadratic Equations by Completing the Square
   
 

Simplifying Sums and Differences of Square Roots

Sums and differences of square roots are often fairly ugly expressions to deal with algebraically, but sometimes we can use the methods for simplification of single square roots to help get lengthier expressions with two or more square root terms into a more compact form. To illustrate ways to simplify a sum or difference of terms containing square roots, consider the following very simple example:

The grouping together of the coefficients 9, 3, -4, in brackets as shown above makes use of the distributive law for multiplication, which we described earlier in these notes in the discussion of the use of brackets.

Alternatively, you could regard

as saying: “we have 9 of these , and another 3 of these , and we’ve taken away 4 of these . That means we have, on balance, (9 + 3 – 4) = 8 of these left.” But “8 of these ” is written mathematically as . (Substitute the word “banana” for the symbol in the sentences above if you’re having trouble understanding the logic of this method.) We can regard this method as collecting together all terms which have identical square root parts.

Now, when the expression involves square roots of different numbers, we first need to simplify each square root as much as possible before collecting the terms with identical square root parts.

 

Example:

Simplify

solution:

We first check the square roots in each of the three terms to see if any of them by themselves can be simplified:

Then

is as simple as possible.

Finally,

Thus, our original expression becomes

This is its simplest form.

 

Example:

Simplify

solution:

Once again, we first simplify each square root part as much as possible:

Thus

This is as simple as we can get the original expression because there is no simple way to combine terms involving with terms involving .

 
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