Solving A Nice Radical Equation
🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts).
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡
/ @sybermath
/ @aplusbi
⭐ Join this channel to get access to perks:→ bit.ly/3cBgfR1
My merch → teespring.com/stores/sybermat...
Follow me → / sybermath
Subscribe → kzread.info?sub...
⭐ Suggest → forms.gle/A5bGhTyZqYw937W58
If you need to post a picture of your solution or idea:
intent/tweet?text...
#algebra #radicals #radicalequations
via @KZread @Apple @Desmos @NotabilityApp @googledocs @canva
PLAYLISTS 🎵 :
▶ Trigonometry: • Trigonometry
▶ Algebra: • Algebra
▶ Complex Numbers: • Complex Numbers
▶ Calculus: • Calculus
▶ Geometry: • Geometry
▶ Sequences And Series: • Sequences And Series
Пікірлер: 14
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
X=(5^(1/2)+1)/2
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.
😊😊😊👍👍👍
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
2 күн бұрын
Nice as always! 😊
Short Video แบบใด 😹 ยาว 9:54 ยาวพอๆ กับ Video ปกติในช่องเลยเธอ
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
error at 4 mn
@renyxadarox
2 күн бұрын
No, the 1st mistake was at 1:27 x+1≥0
@claudelebourlegat
2 күн бұрын
@@renyxadarox no,check till 2mn44s.Moreover my observation is not contradictory to others errors !
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]
(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
x = (√5 + 1) / 2 √(x+1) - √[(x-1)/x] = 1