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MathProwess.
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Can You Solve And Verify This Math Roots?
Can You Solve And Verify This Math Roots?
Watch this video from the beginning to the without skipping any parts for the solution and verification of the roots.
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Watch this video from the beginning to the without skipping any parts for the solution and verification of the roots.
Subscribe my channel for more videos.
www.youtube.com/@UC5f4yCVCO0cz3GlA8WNnV8w
Like, comment and share this video with others.
Thanks a lot.
#maths #mathematics #education #mathstricks #matholympiad
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X=1 is one of the roots.
(x^4 )^2 ➖ (16)^2{x^16 ➖ 256}=240 2^120 2^60 2^30 2^15 2^3^5 2^3^5^1 2^3^1^1 2^3:(x ➖ 3x+2)( x^2)^2 ➖ (4)^2={x^4 ➖ 16}=123^4 3^2^2 1^1^2 1^2 (x ➖ 2x+1) .
Well explained man. Thanks for this detailed work, sir.
Well explained man. Thanks for this detailed work, sir.
❤
Factor to m(x - √3)(x + √3) = 0 then set each factor equal to 0. m = 0 m = +√3 m = -√3 simple
The original looks like m decimal m decimal m on the right side. In that case any non-negative integer will suffice for m.
{x^8+x^8 ➖ }+{z^3+z^3 ➖ }={x^16+z^6}=xz^22 xz^16^6 4^4^3^2 xz^2^2^2^2^3^2 xz^1^1^1^1^3^2 x^3^2(xz ➖ 3xz+2) {y^3+y^3 ➖ }+{x^3+x^3 ➖ }={y^6+x^6}=yx^12 yx^6^6yx^2^3^2^3 yx^2^1^1^3 yx^2^3(yx ➖ 3yx+2) {z^3+z3 ➖ }+{y^3+y^3 ➖ }={z^6+y^6}=zy^12 zy^6^6 zy^2^3^2^3 zy^2^1^1^3 zy^2^3 (zy ➖ 3zy+2).
(x)^3 ➖ (1)^3={x^3 ➖ 1}={x^0+x^0 ➖}=x^1 (x ➖ 1x+1).x^3(x^3)^2 ➖ (3x^2)^2=x^3{x9 ➖ 9x^4}=(x^3 )^2 ➖ 9x^5={x^9 ➖ 9x^5}=9x^4 3^2x^2^2 3^1x^1^2 3x^2 (x ➖ 3x+2)
First to comment...😂😂😂
(2)+(2)=4 (y ➖ 2x+2) 3^2/2 1^1/2 1/2 {xy➖ 2xy+1).
X=887, I just solved it in my head in a few seconds 😅😅😅
👍👍👏👏🙏
X=5, thanks
x^(x)^2 ➖ (2)^2=x^{x^2 ➖ 4}=x^{0+x^0 ➖ }=x^1 (x ➖ 1x+1) 2^22^2 1^1^1^2 1^2 (x ➖ 2x+1).
{x+x+x x ➖ x ➖ x}=x^3 +(x^3 )^2➖( x)^^2 ➖ 16=x^3+ {x^9 ➖ x^2} ➖ 16 =x^3+x^7 ➖ (16)^2=x^3+{x^7 ➖ 256}={x^3+249}=249x^3 10^20^49x^3 10^20^7^7x^3 10^2^10^1^1x^3 2^5^2^2^5x^3 1^1^1^2^1x^3 2x^3(x ➖ 3x+2).(x^3 )^2➖ (x )^2➖ 8={x^9 ➖ x^2 }➖ 8=x^7 ➖ (8)^2={x^7 ➖ 64}= 57 3^19 3^19^1 3^1^1 3^1 (x ➖ 3x+1).
❤🎉
❤❤❤❤❤❤
✅
Thank u
Welcome sir
Good one man.
Thanks for the visit
I don't think this works. (2i)^8=256, and the 4th root of 256 is 4. I mean, if you wanted the roots of x^4=256, you'd have {4, -4, 4i, -4i}, but the radical indicates the principal root, which is 4.
Of course it worked. He did a check.
It works sir. Kindly watch the video clip from the beginning to the end sir. Thanks.
Uncle J.J
Solution overcomplicated! √(7-2√40) =√(7-2√(5·2)) =√(5-2√(5·2)+2)=√((√5-√2)^2 )=√5-√2 or √2-√5
Use the shortcut. √(7 - 2√10), 5 + 2 = 7 and 5 x 2 = 10. Automatic: √5 - √2
1
Not correct sir.
❤
Good 🎉🎉🎉
Thank you! Cheers!
or.. just do sqrt of 16 which is 4 and sqrt of 4 which is 2...
you re missing a square on your cosine
Oops!!!
how is sqrt(cos^4(x)) = |cosx|
🥱🥱🥱🤦🏻♂️🤦🏻♂️And now get the all solutions to x^1024=1... Good luck. LAAAAME
This is a very nice question solved in a clear and unique way.
Thanks sir
I am in 7th grade about to do algebra 2 any tips?
Congrat.
2 real solutions, because -1/e < -ln4/5 < 0 1.5271834618689137843439681976847183206615191189475680905718280703... or 7.0378231590674309036420394728342539963399273768491109730261293405...
No real solutions
Only complex: 0.063959810301343172499210859310985355287605730479104740675403973... - 1.0908433765399575763735988438001669099914583549817516466302112... i or 0.063959810301343172499210859310985355287605730479104740675403973... + 1.0908433765399575763735988438001669099914583549817516466302112... i or 1.248405605048936531526278135104793156575121368698758722353396678... - 5.504574134828021237818776295401377186833156083716398821615796055... i or 1.248405605048936531526278135104793156575121368698758722353396678... + 5.504574134828021237818776295401377186833156083716398821615796055... i or ...
Kudos man...😂😂😢
{4x+4x ➖ }{9x+9x ➖ }={8x^2+18x^2}=26x^4 2^13x^4 2^13^1x^4 2^1^1x^2^2 1x1^2 x^1^2 (x ➖ 2x+1). {6x+6x ➖ }=12x^2 3^4x^2 3^2^2x^2 3^1^1x^2 3x^2 (x ➖ 3x+2).
This beautiful man. Keep running this channel sir and the sky is yours.....❤
Thanks sir
👍👍
Cool one 😅😅
Thanks for watching
Cheap proof.
I pove this in my head....😂😂😂
Pove in your head??
Nice work
Thanks for the visit
Nice proof man.
Thanks!
+1 or -1
That is just two answers out of the 16 roots to this math problem sir, kindly watch the full video for the rest 14th roots which can not be found on the surface. Thanks sir.
Where do you get x_1 = sqrt(2+sqrt(2))/2 + (sqrt(2-sqrt(2))/2 * i from? Same with the other x_(odd number). Not explained. Hint: use the half-angle formula.