Looking at the Difference of Squares

J

Jason76

Senior Member
Joined
Oct 19, 2012
Messages
1,180
Here is another way to look at it. Does this logic sound reasonable? Some books don't talk about taking the 2nd root (square root):

\(\displaystyle 16x^{4} - 9y^{4}\)

\(\displaystyle (\sqrt[2]{16})(\sqrt[2]{x^{4}}) - (\sqrt[2]{9})(\sqrt[2]{y^{4}})\) Taking the 2nd root of the number and variable (respectively) in each term. Easily done with a calculator.

\(\displaystyle (4)(x^{4/2}) - (3)(y^{4/2})\) - This step could be left out if you could quickly recognize the fractional exponent \(\displaystyle 4/2\), and note that it reduces to 2.

\(\displaystyle 4x^{2} - 3y^{2}\)

\(\displaystyle [4x^{2}]^{2} - [3y^{2]^{2}}\) - Stick on a 2nd power on each term.

\(\displaystyle (4x^{2} - 3y^{2})(4x^{2} + 3y^{2})\) - Factor out ;)
 
Last edited:
The goal is for you to recognize (mentally) that 16x^4 is the square of 4x^2 and that 9y^4 is the square of 3y^2.

Once you realize that an expression is one square subtracted from another square, then you ought to move directly to the special factoring pattern for a difference of squares (that you have memorized).

If you need to write out those intermediate steps in order to "see" the given expression as a difference of two square, that's okay for now (eventually, you'll be able to skip all of that), but why post that sqrt(16) or sqrt(9) is found easily with a calculator?

You already know that 4*4=16 and that 3*3=9, so why do you suggest using a calculator to find sqrt(16) and sqrt(9)?

Also, please do not say "second root"; just say square root. :)
 
Last edited:
mmm4444bot said:
The goal is for you to recognize (mentally) that 16x^4 is the square of 4x^2 and that 9y^4 is the square of 3y^2.

Once you realize that an expression is one square subtracted from another square, then you ought to move directly to the special factoring pattern for a difference of squares (that you have memorized).

If you need to write out those intermediate steps in order to "see" the given expression as a difference of two square, that's okay for now (eventually, you'll be able to skip all of that), but why post that sqrt(16) or sqrt(9) is found easily with a calculator?

You already know that 4*4=16 and that 3*3=9, so why do you suggest using a calculator to find sqrt(16) and sqrt(9)?

Also, please do not say "second root"; just say square root. :)
Click to expand...

Again looking at a different way to understand it. The \(\displaystyle 4/2\) exponential fraction needs to be understood in order to be reduced. But then again you could just memorize that the square root of \(\displaystyle x^{4}\) is \(\displaystyle x^{2}\) and also memorize (as in the other problem) that the square root of \(\displaystyle x^{2}\) is \(\displaystyle x\). But knowing the mechanics always helps in an overall understanding of math, but not necessary to solve these problems.
 
Last edited:
Jason76 said:
The \(\displaystyle 4/2\) exponential fraction needs to be understood in order to be reduced.
Click to expand...

I do not agree.

Exponential notation is not required, to get the square root of 16.

Just do it in your head!
 
Jason76 said:
...and also memorize (as in the other problem) that the square root of \(\displaystyle x^{2}\) is \(\displaystyle x\).
Click to expand...
No, the square root of x^2 is \(\displaystyle \ \sqrt{x^2}.\) \(\displaystyle \ \ \ \ \ \ \)\(\displaystyle And \ \ \sqrt{x^2} \ = \ |x|.\)
 
lookagain said:
No, the square root of x^2 is \(\displaystyle \ \sqrt{x^2}.\) \(\displaystyle \ \ \ \ \ \ \)\(\displaystyle And \ \ \sqrt{x^2} \ = \ |x|.\)
Click to expand...

In this expression

\(\displaystyle x^{2} = 16\)

\(\displaystyle \sqrt{x^{2}} = \sqrt{16}\)

\(\displaystyle x = \pm4\)

We see the square root of \(\displaystyle x^{2}\) is \(\displaystyle x\), and the square roots of \(\displaystyle 16\) are positive and negative \(\displaystyle 4\).

What do you mean by absolute value? :confused:
 
Last edited:
Jason76 said:
Here is another way to look at it. Does this logic sound reasonable? Some books don't talk about taking the 2nd root (square root):

\(\displaystyle 16x^{4} - 9y^{4}\)

\(\displaystyle (\sqrt[2]{16})(\sqrt[2]{x^{4}}) - (\sqrt[2]{9})(\sqrt[2]{y^{4}})\) Taking the 2nd root of the number and variable (respectively) in each term. Easily done with a calculator.

\(\displaystyle (4)(x^{4/2}) - (3)(y^{4/2})\) - This step could be left out if you could quickly recognize the fractional exponent \(\displaystyle 4/2\), and note that it reduces to 2.

\(\displaystyle 4x^{2} - 3y^{2}\)

\(\displaystyle [4x^{2}]^{2} - [3y^{2]^{2}}\) - Stick on a 2nd power on each term.

\(\displaystyle (4x^{2} - 3y^{2})(4x^{2} + 3y^{2})\) - Factor out ;)
Click to expand...

That can be further factored in real domain:

\(\displaystyle (2x - y \sqrt{3})(2x + y \sqrt{3})(4x^{2} + 3y^{2})\)
 
Jason76 said:
We see the square root of \(\displaystyle x^{2}\) is \(\displaystyle x\), ...[/tex].
Click to expand...


Jason76, do not ignore what I typed to you. The square root of \(\displaystyle \ x^2 \ = \ \sqrt{x^2} \ = \ |x|.\)


Example: \(\displaystyle \ \sqrt{(-3)^2} \ = \ \sqrt{9} \ = \ 3, \ \ not \ -3, \ \ just \ \ as \ \ |-3| \ = 3.\)
 
lookagain said:
Jason76, do not ignore what I typed to you. The square root of \(\displaystyle \ x^2 \ = \ \sqrt{x^2} \ = \ |x|.\)


Example: \(\displaystyle \ \sqrt{(-3)^2} \ = \ \sqrt{9} \ = \ 3, \ \ not \ -3, \ \ just \ \ as \ \ |-3| \ = 3.\)
Click to expand...

Ok, I got it.
 
You must log in or register to reply here.
Top