Followup: And Substitution is Awesome(ly powerful)
To continue why I think emphasizing substitution early is a good idea see: Previous Post
Here’s a recent more complicated algebraic problem from @_eylem_99:
$x^2 - 16 \sqrt{x} = 12$. What is $x - 2\sqrt{x}$?
[And I just realized I’ve talked about this one before: not-so-innocuous-quartic]
As usual I do a quick substitution let $ y = \sqrt{x} $ to get a better picture of the equation (and make it easier to manipulate) and get $ y^4 - 16y - 12 = 0 $ So this is really a quartic in disguise where you want to find $y^2 - 2y $.
Brute Force Attack My first instinct is to always factor if you can: in this case there is no easy rational root but it must be the product of two quadratics and it quickly becomes apparent that they must be semi complementary i.e. $(y^2 - ay + b)(y^2 + ay - c) = 0 $ This leads to the further system:
$b + c = a^2$
$ab - ac = -16 $
$bc = -12 $
Which is not terribly easy to solve although a factoring approach for the constant term can lead eventually to: $ (y^2 + 2y - 2)(y^2 - 2y + 6) = 0$ At which point the first factor/root contains the solution if $y^2 + 2y - 2 = 0 $ then $y^2 + 2y = 2$ Overall this is fairly hairy - see the post lined above for more details about the formal method
Clever Factoring A second approach relies on clever factoring and trying to complete the square:
1. $ x^2 = 16 \sqrt{x} + 12 $
2. Add 4x + 4 : $ x^2 + 4x + 4 = 16 \sqrt{x} + 4x + 16 $
3. $(x + 2)^2 = (2\sqrt{x} + 4)^2 $
From there its easy to take a square root and solve for x to get $ x - 2 \sqrt{x} = 2 $ but finding that exact extra term to create a square on both sides is an** exercise in intuition and experimentation.**
Substitution I think a semi recursive substitution provides a more natural solution i.e. less need for an intuitive leap:
1. Move everything to the side as above: $ x^2 = 16 \sqrt{x} + 12 $
2. let $x - 2\sqrt{x} = k $ then $ x = k + 2\sqrt{x} $
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Square that term: $ x^2 = k^2 + 4k\sqrt{x} + 4x $
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And then substitute for x again: $ x^2 = k^2 + 4k\sqrt{x} + 4(k + 2\sqrt{x}) $
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Setting that equal to the original equation $ k^2 + 4k = 12 $ and $ 4k + 8 = 16 $. The easier linear equation implies k = 2 and that is consistent with the quadratic so indeed: $x - 2\sqrt{x} = 2 $
[Addon]
https://en.wikipedia.org/wiki/Reciprocal_polynomial
This all follows once you substitute y = 1/x in a polynomial but its still surprisingly beautiful.
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