Russia | A nice Math Olympiad Algebra Problem | Quartic Simplification | Can you solve this ?
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- Опубликовано: 23 июн 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!!
Go for (✓3 - 1)
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
As another way,
let the given E, then
E =⁴√(28 -16√3)
=⁴√(28 -2•8√3)
=⁴√(28 -2√192)
=⁴√[28 -2(√16)(√12)]
=⁴√[16 +12 -2(√16)(√12)]
=⁴√(√16 -√12)²
that is,
E =√(√16 -√12)
=√(4 -2√3)
=√(3 +1 -2√3)
=√(√3 +1 -2√3)
=√(√3 -1)²
Therefore,
E = √3 -1
I tried but don't know what was missed.
Actually is sqrt(3) -1
Please re-check!
Your method is interesting but I used a different method and got a different answer. Here's what I did:
(28 - 16√3)^.25 This looked suspiciously like many √(a - 2√b) problems on RUclips except that it was to the ¼ power. So I broke it into two nested square roots.
( (28 - 16√3)^.5 )^.5
( (28 - 2√192)^.5 )^.5
( (28 - 2√(16*12))^.5 )^.5 The inner square root is of the form √(a^2 + b^2 + 2ab), which equals a + b. Here a = √16 and b = √12.
(√16 - √12 )^.5
( 4 - 2√3 )^.5
( 4 - 2√(3*1) )^.5 Do the same thing again.
√3 - √1
√3 - 1
I checked the result with a calculator.
16√3 = 27.713
28 - 16√3 = .287
.287^¼ = .732 which equals √3 - 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
You said that x²+y²
The answer is √3-1(=0.732), not √3+1.
(28-16√3)^(1/4)=[(√16-√12)^2]^(1/4)=(√16-√12)^(1/2)=[(√3-1)^2]^(1/2)=√3-1.
答案是根號3-1，非根號3+1
13:46 u missed x=sqrt(3)-1 😅
28_16 кв.кор3=(4_2квкор3)в кв=квкор(квкор3_1)в кв=(квкор3_1)в 4степ=квкор3_1
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)
Soluția este greșita radical din 3-1 e buna
答案是根號3-1
Have a error 3^(1/2)-1
Wrong answer, correct result must be sqrt(3)-1 in stead of sqrt(3)+1.
16 + 12 - 2.4.√12 = (4-√12)^2= ( 3 +1 - 2√3)^2 = (√3-1)^4
X = √3-1
（√3₊1)^4＝(4+2√3)^2=28+16√3=y^4 ∴y=√3+1
（√3 -1)^4＝(4 -2√3)^2=28 -16√3=x^4 ∴x=√3 -1
There are easier ways to get the solution:
∜(28 - 16 √3) = ∜(16 + 12 - 2∙4∙2√3) = ∜(4² + (2√3)² - 2∙4∙2√3) = ∜((4 - 2√3)²) = √(4 - 2√3)
√(4 - 2√3) = √(√3² + 1² - 2√3) = √(√3 - 1)²) = √3 - 1
btw: The final solution in the video is wrong. The correct answer is √3 - 1 and not √3 + 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=2√3
x+(2/x)=2√3
x²-(2√3)x+2=0
∆=(2√3)²-4(1)(2)=12-8=4
x=(-(-2√3)±√4)/2=(2√3±2)/2=√3±1
because x>0 => x=√3-1
because you accept x+y=-2√3 ,you are solve equation: x²+(2√3)x+2=0
∆=(2√3)²-4(1)(2)=4
x=(-2√3±√4)/2=-√3±1
because x>0 => x=√3+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