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2_sq_roots_p2
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We have

**A:**

and we want to get

**B:**

1 ---------- end topic
The inequalities in **A** are of the form , which imply and give us the inequality in **B**.

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*If* one has studied *and* mastered algebra, then the equality in **B** is immediately evident: The square of a two term sum is the sum of the squares of the the two terms *plus* two times their product. The square of the difference of two terms is the sum of the squares of the two terms *minus* two times their product**,**

and when these are subtracted we get 4*ax***.** However, if one has no mastery of algebra but *has* studied and mastered the principles of the real number system, then algebra - and arithmetic - can be worked out from scratch. In the present problem, we might proceed as follows**:**

By the definition of integral exponents we have

which by the distributive property is equal to

which by the commutative property is equal to

which by the distributive property is equal to

which by the definition of integral exponents and commutativity is equal to

which by the neutral element for multiplication is equal to

which by the commutative property is equal to

which by the distributive property is equal to

which by the definition of 2 is equal to

which by commutativity is equal to

Then, by the definition of subtraction we have

which has the same form as the case treated above, and is therefore equal to

and now consider what happens as we gain experience with these principles over time. We begin after a while to recognize situations which occur repeatedly and where a certain principal is applicable. We look at the last term, (-*a*) squared, and we think as follows without any need to write anything on paper:

Minus *a* is equal to minus 1 times *a* ... the square of a product is the product of the squares ... negative times negative is positive ... so we can replace minus *a* squared by *a* squared.

In the second term we have two times *x* times minus one times *a* ... we can put the minus one first with the two the *x* and the *a* in parenthesis representing one number ... then minus one times that number is just minus that number ... then we use the definition of subtraction.

So we can write down the result immediately to get

and by subtraction the final result is

4*xa***.**

With practice, these immediately recognized situations become more numerous and complex.