Russia | A nice Math Olympiad Algebra Problem | Quartic Simplification | Can you solve this ?
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- Äas pĆidĂĄn 23. 06. 2024
- Russia | A nice Math Olympiad Algebra Problem | Quartic Simplification | Can you solve this ?
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Mathematical thought. The square root and squaring do NOT cancel.!!The square root of x squared is the absolute value of X.Thats where the plus or minus come from!!
As another way,
let the given E, then
E =âŽâ(28 -16â3)
=âŽâ(28 -2âą8â3)
=âŽâ(28 -2â192)
=âŽâ[28 -2(â16)(â12)]
=âŽâ[16 +12 -2(â16)(â12)]
=âŽâ(â16 -â12)ÂČ
that is,
E =â(â16 -â12)
=â(4 -2â3)
=â(3 +1 -2â3)
=â(â3 +1 -2â3)
=â(â3 -1)ÂČ
Therefore,
E = â3 -1
I tried but don't know what was missed.
Your method is interesting but I used a different method and got a different answer. Here's what I did:
(28 - 16â3)^.25 This looked suspiciously like many â(a - 2âb) problems on CZcams except that it was to the ÂŒ power. So I broke it into two nested square roots.
( (28 - 16â3)^.5 )^.5
( (28 - 2â192)^.5 )^.5
( (28 - 2â(16*12))^.5 )^.5 The inner square root is of the form â(a^2 + b^2 + 2ab), which equals a + b. Here a = â16 and b = â12.
(â16 - â12 )^.5
( 4 - 2â3 )^.5
( 4 - 2â(3*1) )^.5 Do the same thing again.
â3 - â1
â3 - 1
I checked the result with a calculator.
16â3 = 27.713
28 - 16â3 = .287
.287^ÂŒ = .732 which equals â3 - 1
I haven't gone through yours in detail but it seems that +'s and -'s on the second term got mixed up somewhere.
Go through your method again to correct the anomaly
x and y are both positive (from the definitions), so a lot of the +- arguments are not needed. The issue is solving equations 2 and 3 together... do the same for y and you get the same quadratic, with the same 2 answers ie SQRT(*3) +- 1... the author has just picked the wrong one for x... y>x so one is SQRT(3) + 1 --> y and the other SQRT(3) - 1 is x
Please re-check!
28-16sqrt(3)-=(2sqrt(3))^2+4^2-2(4)(2sqrt(3))=(4-2sqrt(3))^2. so (28-16sqrt(2))^(1/4)= sqrt(4-2sqrt(3)). Now 4-2sqrt(3)= 1^2+(sqrt(3))^2-2(1)(sqrt(2))=(sqrt(3)-1)^2. So sqrt(4-2sqrt(3))=sqrt(3)-1 or 1-sqrt(3)
28_16 ĐșĐČ.ĐșĐŸŃ3=(4_2ĐșĐČĐșĐŸŃ3)ĐČ ĐșĐČ=ĐșĐČĐșĐŸŃ(ĐșĐČĐșĐŸŃ3_1)ĐČ ĐșĐČ=(ĐșĐČĐșĐŸŃ3_1)ĐČ 4ŃŃДп=ĐșĐČĐșĐŸŃ3_1
ïŒâ3â1)^4ïŒ(4+2â3)^2=28+16â3=y^4 âŽy=â3+1 ăă
ïŒâ3 -1)^4ïŒ(4 -2â3)^2=28 -16â3=x^4 âŽx=â3 -1
16 + 12 - 2.4.â12 = (4-â12)^2= ( 3 +1 - 2â3)^2 = (â3-1)^4
X = â3-1
The answer is â3-1(=0.732), not â3+1.
(28-16â3)^(1/4)=[(â16-â12)^2]^(1/4)=(â16-â12)^(1/2)=[(â3-1)^2]^(1/2)=â3-1.
çæĄæŻæ čè3-1ïŒéæ čè3+1
SoluÈia este greÈita radical din 3-1 e buna
çæĄæŻæ čè3-1
Have a error 3^(1/2)-1
You said that xÂČ+yÂČ
There are easier ways to get the solution:
â(28 - 16 â3) = â(16 + 12 - 2â4â2â3) = â(4ÂČ + (2â3)ÂČ - 2â4â2â3) = â((4 - 2â3)ÂČ) = â(4 - 2â3)
â(4 - 2â3) = â(â3ÂČ + 1ÂČ - 2â3) = â(â3 - 1)ÂČ) = â3 - 1
btw: The final solution in the video is wrong. The correct answer is â3 - 1 and not â3 + 1.
What a waste!
Is the concept of denesting a waste of time?
@@superacademy247 At least you should pin up a message for viewers in the future explaining why the result is wrong so they don't get confused.
x>0 and y>0 so x+y>0
in 10:42 because x+y>0 => x+y=2â3
x+(2/x)=2â3
xÂČ-(2â3)x+2=0
â=(2â3)ÂČ-4(1)(2)=12-8=4
x=(-(-2â3)±â4)/2=(2â3±2)/2=â3±1
because x>0 => x=â3-1
because you accept x+y=-2â3 ,you are solve equation: xÂČ+(2â3)x+2=0
â=(2â3)ÂČ-4(1)(2)=4
x=(-2â3±â4)/2=-â3±1
because x>0 => x=â3+1 and this is wrong answer
I made a mistake because there are four solutions in which out of it one is valid. Thanks for your concern