Can You Solve This Algebra Problem WITHOUT SQUARING BOTH SIDES? Math Olympiad Mathematics
NonsoMaths
3,742 views
In this math tutorial, we solve the equation x³-3x = √(x+2) using a breathtakingly elegant trigonometric substitution that transforms both sides into standard cosine expressions simultaneously. We substitute x = 2cos(θ), which converts the left hand side using the triple angle identity cos(3θ) = 4cos³(θ)-3cos(θ) to give 2cos(3θ), and converts x+2 = 2cos²(θ/2)·2 using the half angle identity 1+cos(θ) = 2cos²(θ/2), so √(x+2) = 2cos(θ/2). The equation reduces to 2cos(3θ) = 2cos(θ/2), and hence cos(3θ) = cos(θ/2). We solve this trigonometric equation by considering the two cases 3θ = ±θ/2+2kπ, find all valid values of θ in the domain, and compute x = 2cos(θ) for each. We verify all solutions and apply domain restrictions. Every step is explained clearly from start to finish.
The substitution x = 2cos(θ) is the entire key — it simultaneously activates the triple angle identity on the left and the half angle identity on the right, converting an algebraic equation that seems to mix two completely different types of expressions into a single clean trigonometric equation.
What you will learn:
How x=2cos(θ) activates the triple angle identity on x³-3x
How 2+x=2+2cos(θ)=4cos²(θ/2) activates the half angle identity on √(x+2)
How the equation reduces to cos(3θ)=cos(θ/2)
How to solve cos(3θ)=cos(θ/2) using both cases 3θ=θ/2+2kπ and 3θ=-θ/2+2kπ
How to find all valid values of θ and compute x=2cos(θ)
How to verify all solutions and apply domain restrictions x≥-2
This type of trigonometric substitution solving a cubic radical equation appears in the most prestigious math olympiad competitions, international mathematics contests, IB Mathematics Higher Level, A-Level Further Mathematics, and university entrance examinations. The simultaneous activation of triple and half angle identities by a single substitution is one of the most beautiful techniques in all of competitive mathematics.
If this gave you that rare feeling where trigonometry and algebra meet in a completely unexpected and deeply satisfying way, give it a thumbs up, share it with someone who loves beautiful mathematics, and subscribe for weekly videos on algebra, number theory, calculus, and olympiad problem solving.
Hit the notification bell. The next problem might surprise you even more.
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#maths #algebra #exam
9 Comments
x^3 - 3x = √(x + 2)
At x = 2 we have 8 - 6 = √(2 + 2) or 2 = 2
We can substitute so that we don’t square to try to simplify:
let y = √(x + 2)
y^2 = x + 2
x = y^2 - 2
x^3 - 3x = y
x = y^2 - 2
x^3 - 4x = y - y^2 + 2
x^3 - 4x + y^2 - y - 2 = 0
x(x + 2)(x - 2) + (y + 1)(y - 2) = 0
y - 2 = √(x + 2) - 2 = ((√(x + 2) - 2)(√(x + 2) + 2))/(√(x + 2) + 2) = (x + 2 - 4)/(y + 2) = (x - 2)/(y + 2)
x(x + 2)(x - 2) + (y + 1)((x - 2)/(y + 2)) = 0
x(x + 2)(x - 2) + ((y + 1)/(y + 2))(x - 2) = 0
using x = 2 from earlier we can take x = 2 out now with some manipulation shown above.
(x - 2)(x^2 + 2x + (y + 1)/(y + 2)) = 9
x = 2
x^2 + 2x + (y + 1)/(y + 2) = 0
x = (-2 ± √(4 - ((4y + 4)/(y + 2))/2
The part inside the square root looks nasty but it’s not really in the numerator because it becomes this:
(4y + 8 - 4y - 4)/(y + 2)
So x = (-2 ± √(4/(y + 2))/2
x = -1 ± √(1/(y + 2)
And that’s still horrible so
x(x + 2) + (√(x + 2) + 1)/(√(x + 2) + 2) = 0
The other solutions without going into it much further are x ≈ - 0.455 or 2 cos (4π/7) and x = -φ or (-1 - √5)/2 while the integer solution remains x = 2.
The other way of writing the other solutions within actually giving them is provided with 1/(y + 2) inside the radical but then to extract x you wind up with a 4th degree polynomial and 2 erroneous solutions eliminating the 5th degree polynomial at least
Can You Solve This Algebra Problem WITHOUT SQUARING BOTH SIDES? Math Olympiad Mathematics
NonsoMaths
3,742 views
In this math tutorial, we solve the equation x³-3x = √(x+2) using a breathtakingly elegant trigonometric substitution that transforms both sides into standard cosine expressions simultaneously. We substitute x = 2cos(θ), which converts the left hand side using the triple angle identity cos(3θ) = 4cos³(θ)-3cos(θ) to give 2cos(3θ), and converts x+2 = 2cos²(θ/2)·2 using the half angle identity 1+cos(θ) = 2cos²(θ/2), so √(x+2) = 2cos(θ/2). The equation reduces to 2cos(3θ) = 2cos(θ/2), and hence cos(3θ) = cos(θ/2). We solve this trigonometric equation by considering the two cases 3θ = ±θ/2+2kπ, find all valid values of θ in the domain, and compute x = 2cos(θ) for each. We verify all solutions and apply domain restrictions. Every step is explained clearly from start to finish.
The substitution x = 2cos(θ) is the entire key — it simultaneously activates the triple angle identity on the left and the half angle identity on the right, converting an algebraic equation that seems to mix two completely different types of expressions into a single clean trigonometric equation.
What you will learn:
How x=2cos(θ) activates the triple angle identity on x³-3x
How 2+x=2+2cos(θ)=4cos²(θ/2) activates the half angle identity on √(x+2)
How the equation reduces to cos(3θ)=cos(θ/2)
How to solve cos(3θ)=cos(θ/2) using both cases 3θ=θ/2+2kπ and 3θ=-θ/2+2kπ
How to find all valid values of θ and compute x=2cos(θ)
How to verify all solutions and apply domain restrictions x≥-2
This type of trigonometric substitution solving a cubic radical equation appears in the most prestigious math olympiad competitions, international mathematics contests, IB Mathematics Higher Level, A-Level Further Mathematics, and university entrance examinations. The simultaneous activation of triple and half angle identities by a single substitution is one of the most beautiful techniques in all of competitive mathematics.
If this gave you that rare feeling where trigonometry and algebra meet in a completely unexpected and deeply satisfying way, give it a thumbs up, share it with someone who loves beautiful mathematics, and subscribe for weekly videos on algebra, number theory, calculus, and olympiad problem solving.
Hit the notification bell. The next problem might surprise you even more.
Don’t forget to like 👍, subscribe @nonsomaths, and hit the notification bell for more math tips and tricks!
#maths #algebra #exam
k=0; @=0; X=2;
k=1; @=4π/5; X=(-1-√5)/2;
@=4π/7; X=2cos(4π/7)~-0.44507