Solving A Nice Radical Equation

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[@Foamea45]

3:58 You get x^2 at the denominator in the left side and x in the right side.the denominators don't cancel,it will remain an x in the left side

[@epimaths]

good math....

[@MatteressZ]

Hi, great video but I got a question: Why doesn't our inital domain restriction help us remove any extraneous solutions at the end? eg. (-1,0) or (1, inf) and 1/2-sqrt5/2 is roughly -0.61 which is between -1 and 0.

[@phill3986]

😊😊😊👍👍👍

[@StaR-uw3dc]

Substitution t=√(x+1)≥0 leads to quartic t⁴-2t³-t²+2t+1 = (t²-t-1)² = 0 i.e. t= (1+√5)/2 and x=t²-1=(1+√5)/2. Substitution y=√((x-1)/x)≥0 leads to quartic y⁴+2y³-y²-2y+1= (y²+y-1)² = 0 i.e. y=(-1+√5)/2 and x= 1/(1-y²) = (1+√5)/2

[@JidtapadTK]

Short Video แบบใด 😹 ยาว 9:54 ยาวพอๆ กับ Video ปกติในช่องเลยเธอ

[@yakupbuyankara5903]

X=(5^(1/2)+1)/2

[@claudelebourlegat]

error at 4 mn

[@renyxadarox]

No, the 1st mistake was at 1:27 x+1≥0

[@claudelebourlegat]

@@renyxadarox no,check till 2mn44s.Moreover my observation is not contradictory to others errors !

[@SidneiMV]

x + 1 + (x - 1)/x - 2√[(x² - 1)/x] = 1 2√[(x² - 1)/x] = (x² + x - 1)/x 4(x² - 1)/x = (x² + x - 1)²/x² 4x(x² - 1) = (x² + x - 1)² (x² - 1)² + x² + 2x(x² - 1) - 4x(x² - 1) = 0 (x² - 1)² + x² - 2x(x² - 1) = 0 (x² - x - 1)² = 0 x² - x - 1 = 0 *x = (1 + √5)/2* [by verification (1 - √5)/2 is not valid]

[@stratosleounakis2267]

(Χ+1)^1/2 - [(1-1/Χ]^1/2 =1 1/Χ = Y [(Χ+1)^1/2 - (1-Y)^1/2]^2 =1^2 Χ+1+1-Y-2[(Χ+1)(1-Y]^1/2 =1 Χ+2-Y-2(Χ-ΧY+1-Y)^1/2 = 1, ΧY =1 Χ-Y+2 -2(Χ-Y)^1/2 = 1 Χ-Y-2(Χ-Y)^1/2+1= Ο, (Χ-Ψ)^1/2 = Α Α^2-2Α+1= 0 (Α-1)^2 = 0 Α-1 = 0, A^2 = 1 , X-Y = 1, Y = 1/X X-1/X = 1 X^2-X-1 = 0 X = [1+ (5)^1/2] / 2

[@DonEnsley-mathdrum]

x = (√5 + 1) / 2 √(x+1) - √[(x-1)/x] = 1

[@MrGeorge1896]

Already guessed the golden ratio before solving. But don't ask me why 😁 btw I started with the quartic but saw the coefficients being symmetrical so I divided the whole thing by x² and ended up the same way as in the video. (x² - x - 1 = 0)

[@ShortsOfSyber]

Nice as always! 😊