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Thought-Process to Discover Knowledge

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Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

Read in the blogs more about the unique learning experience at nubtrek.continue

In this page, standard algebraic identities with exponent 2 are introduced and explained.

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What does 'identity' mean?

  • It is just a name.
  • equality of two expressions; left and right hand side are identical
  • equality of two expressions; left and right hand side are identical

Answer is 'equality of two expressions; left and right hand side are identical'

Identities are statements that specify two expressions as equal.

In numerical terms, some of the identities are
 •  `5xx(2+3) = 5xx2+5xx3`
(left hand side equals to right hand side )
 •  `4-0 = 4`
(left hand side and right hand side are equal.)

Similar to identities of numerical expressions, algebraic identities are statements that specify two algebraic expressions as equal.
(left hand side equals right hand side)
 •  `a^2+2a^2+b = 3a^2+b`
 •  `(2+x)x = x^2+2x`

Some identities are repeatedly used. Such identities are called standard algebraic identities. While handling algebraic expressions, it helps to identify the left hand side or right hand sides of the standard identities.

Students are taught with these standard algebraic identities.

Which of the following equals `2xx(4+1)`?

  • `2xx4 + 1`
  • `2xx4 + 2 xx1`
  • `2xx4 + 2 xx1`

The answer is '`2xx4 + 2 xx1`'.

This is verified by finding the value of the numerical expressions

Question: `2xx(4+1)``=2 xx5``=10`

first choice: `2xx4 + 1` `=8+1``=9`

second choice: `2xx4+2xx1` `=8+2``=10`

The expression given in question equals the second choice.

Generalizing the numbers to variables.
`axx(b+c) = axxb+axxc`.

Note that `a` is a measure of quantity given by a variable.
The variable `a` can be `4` or `2+2` or `3+1` or `(p+q)`.
This is given as
`color(coral)(a)xx color(deepskyblue)((b+c)) ``= color(coral)(a)xx color(deepskyblue)b+color(coral)(a)xx color(deepskyblue)c`

`color(coral)(4)xx color(deepskyblue)((b+c)) ``= color(coral)(4)xx color(deepskyblue)b+color(coral)(4)xx color(deepskyblue)c`

`color(coral)((2+2))xx color(deepskyblue)((b+c)) ``= color(coral)((2+2))xx color(deepskyblue)b+color(coral)((2+2))xx color(deepskyblue)c`

`color(coral)((3+1))xx color(deepskyblue)((b+c)) ``= color(coral)((3+1))xx color(deepskyblue)b+color(coral)((3+1))xx color(deepskyblue)c`

`color(coral)((p+q))xx color(deepskyblue)((b+c)) ``= color(coral)((p+q))xx color(deepskyblue)b + color(coral)((p+q)))xx color(deepskyblue)c`

We have understood that
`color(coral)((p+q)) xx color(deepskyblue)((b+c)) ``= color(coral)((p+q))xxcolor(deepskyblue)(b) + color(coral)((p+q))xxcolor(deepskyblue)(c)`

The same proof can be used to take in `color(deepskyblue)b` and `color(deepskyblue)(c)` into `color(coral)((p+q))`. This is given below.

`color(coral)((p+q))xx color(deepskyblue)((b+c))`
`= color(coral)((p+q))xxcolor(deepskyblue)(b) + color(coral)((p+q))xxcolor(deepskyblue)(c)`
`= color(coral)(p)xx color(deepskyblue)(b) + color(coral)(q)xxcolor(deepskyblue)(b) + color(coral)(p)xx color(deepskyblue)(c) + color(coral)(q)xxcolor(deepskyblue)(c)`

This is a very important result in understanding identities.
It is important for students to understand how this result is derived.

`color(coral)((p+q))xx color(deepskyblue)((b+c))`
`= color(coral)(p) color(deepskyblue)(b) + color(coral)(q)color(deepskyblue)(b) + color(coral)(p) color(deepskyblue)(c) + color(coral)(q)color(deepskyblue)(c)`

All algebraic identities use this result to prove that left-hand-side equals right-hand-side.

Do not memorize any identities by rote. Derive them quickly on the fly using this result.
With repeated use and practice, one will remember.

Which of the following is equivalent to `(a+b)^2`?

  • `(a+b)(a+b)`
  • `(a+b)(a+b)`
  • `a+b^2`
  • `a^2+b^2`

The answer is '`(a+b)(a+b)`'


`= color(coral)((a+b))color(deepskyblue)((a+b))`

`= color(coral)((a+b))color(deepskyblue)(a) ``+ color(coral)((a+b))color(deepskyblue)(b)`

`= color(coral)(a)color(deepskyblue)(a) + color(coral)(b)color(deepskyblue)(a) ``+ color(coral)(a)color(deepskyblue)(b)``+ color(coral)(b)color(deepskyblue)(b)`

`= a^2+ 2ab + b^2`

Which of the following is equivalent to `(a-b)^2`?

  • `(a-b)(a-b)`
  • `(a-b)(a-b)`
  • `a-b^2`
  • `a^2-b^2`

The answer is '`(a-b)(a-b)`'


`= color(coral)((a-b))color(deepskyblue)((a-b))`

`= color(coral)((a-b))color(deepskyblue)(a) ``- color(coral)((a-b))color(deepskyblue)(b)`

`= color(coral)(a)color(deepskyblue)(a) - color(coral)(b)color(deepskyblue)(a) ``- color(coral)(a)color(deepskyblue)(b)``-(- color(coral)(b)color(deepskyblue)(b))`

`= a^2 - 2ab + b^2`

Comparing the two identities

`(a_1+b_2)^2 = a_1^2+2a_1b_1+b_1^2`
`(a_2-b_2)^2 = a_2^2-2a_2b_2+b_2^2`

Substituting in the first identity `b_1 =- b_2`; the second identity is derived. Students may work this out to understand.

Which of the following is equivalent to `(a+b)(a-b)`?

  • `a(a-b)+b(a-b)`
  • `aa-ba+ba-b b`
  • `a^2-b^2`
  • all the above
  • all the above

The answer is 'all the above'


`= color(coral)((a+b))color(deepskyblue)(a) ``- color(coral)((a+b))color(deepskyblue)(b)`

`= color(coral)(a)color(deepskyblue)(a) + color(coral)(b)color(deepskyblue)(a) ``- color(coral)(a)color(deepskyblue)(b)``- color(coral)(b)color(deepskyblue)(b)`

`= a^2 - b^2`

Which of the following is equivalent to `(x+a)(x+b)`?

  • `x+ax+b`
  • `x+ab`
  • `x^2+(a+b)x+ab`
  • `x^2+(a+b)x+ab`

The answer is '`x^2+(a+b)x+ab`'


`= color(coral)((x+a))color(deepskyblue)(x) ``+ color(coral)((x+a))color(deepskyblue)(b)`

`= color(coral)(x)color(deepskyblue)(x) + color(coral)(a)color(deepskyblue)(x) ``+ color(coral)(x)color(deepskyblue)(b)``+ color(coral)(a)color(deepskyblue)(b)`

`= x^2+ (a+b)x + ab`

Comparing the two identities

`(x+a_1)(x+b_1) ``= x^2+ (a_1+b_1)x + a_1b_1`

`(a_2+b_2)^2 ``= a_2^2+2a_2b_2+b_2^2`

Substituting in the first identity `x=a_2` ; `a_1=b_2`; and `b_1 = b_2`; the second identity is derived. Students may work this out to understand.

It is known that `(a+b)^2 = a^2+2ab+b^2`, can this result be used to find `(a+b+c)^2`?

  • Yes. Consider `a+b+c = a+ (b+c)` where `b+c` is a single number
  • Yes. Consider `a+b+c = a+ (b+c)` where `b+c` is a single number
  • No. It is a new identity

The answer is 'Yes. Consider `a+b+c = a+ (b+c)` where `b+c` is a single number'

`(color(coral)(a+)color(deepskyblue)((b+c))) ^2`

`= color(coral)(a^2)+ 2color(coral)(a)color(deepskyblue)((b+c)) ``+ color(deepskyblue)((b+c)^2)`

`= color(coral)(a^2) + 2color(coral)(a)color(deepskyblue)(b) ``+ 2color(coral)(a)color(deepskyblue)(c)``+ color(deepskyblue)(b^2+2bc+c^2)`

`= a^2+b^2+c^2+2ab+2bc+2ca`

algebraic identities are derived by expanding the multiplication in algebraic expressions.

Algebraic Identities of Square :

`(x+y)^2 ``= x^2+2xy+y^2`

`(x-y)^2 ``= x^2-2xy+y^2`

`(x+y)(x-y) ``= x^2-y^2`

`(x+a)(x+b) ``= x^2+(a+b)x+ab`

`(x+y+z)^2 ``= x^2+y^2+z^2``+2xy+2yz+2zx`

Solved Exercise Problem:

What is `(a+b-c)^2`?

  • `a^2+2a(b-c)+(b-c)^2`
  • `a^2+b^2+c^2+2ab``+2bxx(-c)+2xx(-c)xxa`
  • `a^2+b^2+c^2+2ab-2bc-2ca`
  • all the above
  • all the above

The answer is 'all the above'

slide-show version coming soon