An Amazing Algebra Challenge | Can You Solve?

An Amazing Algebra Challenge | Can You Solve?
Welcome to InfyGyan!
Join us in the quartic equation challenge, where we watch quick and efficient solution using algebraic formula's with equation solving tactics! 🚀 Whether you're a student, enthusiast, or problem-solving aficionado, this video breaks down quartic equations in a way that's easy to grasp and apply.
🧠💡 Get ready to conquer the challenge and boost your equation-solving skills!
📚 Topics covered:
Introduction to Quartic Equations
Quick Techniques (identities, substitutions and manipulations) for Solutions
Converting into Quadratic equation
Completing the squares method, complex number
Pro tips for Success
🎓 Level: Beginner to Intermediate
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Additional resources:
• Solving Amazing Quarti...
• How to CRACK This Quar...
• Want To Ace Quartics? ...
• The Easy Way to Solve ...
#quarticequations #canyousolve #matholympiad #problemsolving #mathskills #education #learnmaths #quadraticequations #mathtutorial #algebra #math
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🎓 Happy learning, and see you in the next video! 🎉
Thank you for watching video.

[@gregevgeni1864]

Another approach
Let x = √2•t (*).
The given equation is rewritten as
(√2•t)⁴+(√2•t+√2)⁴=68
4t⁴+4(t+1)⁴=68 t⁴+(t+1)⁴=17
t⁴+2t³+3t²+2t-8=0
(t-1)(t+2)(t²+t+4)=0 by inspection.
t=1 or t=-2 =>
x=√2 or x=-2√2 due to (*).

[@davidseed2939]

here a different method,
substitute u=x/√2
u^4+(u+1)^4=2^4+1
soln 1
then make u^4=2^4 and (u+1)^4=1 u=-2
and u^4=1 (u+1)^4=2^4 u=1

[@潘博宇-k4l]

X=(-2√2，√2)

[@RashmiRay-c1y]

Let y=x+√2. Then x^4+y^4=68 and x-y=-√2. Let xy=a. Then, x^4+y^4=2a^2+8a+4=68 which gives a=-8,4. x=-8 does not yield real solutions. For a=4, we get, from x-y=-√2 and xy = 4, x=√2, -2√2.

[@gregevgeni1864]

x = √2 , x = -2√2 real solutions.
The given equation is equivalent to
x⁴+2√2x³+6x²+4√2x-32=0
x³(x+2√2)+2(3x²+2√2x-16)=0
x³(x+2√2)+2[3x(x+2√2)-4√2(x+2√2)=0 (x+2√2)(x³+6x-8√2)=0
x+2√2=0 or x³+6x-8√2=0 =>
x=-2√2 or x=√2.
x³+6x-8√2=x³-2√2+6x-6√2=
x³-(√2)³+6(x-√2)=
(x-√2)(x²+√2•x+√2²)+6(x-√2)=
(x-√2)(x²+√2•x+8)=0 => x=√2 etc . .

[@SidneiMV]

x + (√2)/2 = u => x = u - (√2)/2
[u - (√2)/2]⁴ + [u + (√2)/2]⁴ = 68
2(u⁴ + 3u² + 1/4) = 68
u⁴ + 3u² - 135/4 = 0
u² = (-3 + 12)/2
u² = 9/2 => u = ±(3√2)/2
u = (3√2)/2
x = (3√2)/2 - (√2)/2
*x = √2*
u = (-3√2)/2
x = (-3√2)/2 - (√2)/2
*x = -2√2*

[@Fjfurufjdfjd]

Η λυση στη φωτο. (βαριεμαι να ξαναγραφω)

[@RealQinnMalloryu4]

x^4 +(x^4+20)={x^4+x^4 ➖ }+20x^4={x^8+20x^4}=20x^12 4^4^4^4^4x^4^8 2^2^2^2^2^2^2^2^2^2x^2^2^2^3 1^1^1^1^1^1^1^1^1^1x^1^1^1^1^2 1x^1^2 1x^2 (x ➖ 2x+1).

[@SH1234]

for Quick solution:
x⁴+(x+√2)⁴=68=(4+64) or (64+4); =>x⁴=4 or 64;
=> x⁴=(√2)⁴ or (2√2)⁴;=>
x = ±√2; or; ±2√2; verify>
for x =±√2; (x+√2)⁴=64; => x +√2=2√2;=>x=√2;
x=-√2 not verified >rej.
for x=±2√2 =>x=-2√2 is
Verified, & 2√2 not ver.
x = √2 or -2√2