Let's Solve A Special Radical Equation

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[@allykid4720]

a^6 = a + 62; where a^2 = x 0 = 62 (mod a); Implying: a =+-31, or a = +- 2; and a =+-62, or a = +- 1; 31^6 and 62^6 are too much, 1 is not a solution either, but 2^6 = 64 = 2+62. So: x = 2^2 = 4.

[@MattHudsonAtx]

Oh my fuck now I can do modular math

[@allanmarder456]

The "3rd method" is not really much of an improvement. It's just as easy guessing x=4 in the original equation as it is guessing y=2 in the modified equation. Also factoring y^6 -y -62 is not very helpful.... y^6 -y -62 factors as (y-2)[y^5 + 2*(y^4) + 4*(y^3) +8*(y^2) +16*y +31]. The second factor as only 1 real solution. Using Newton-Raphson that root is approximately y = -1.978719 which is extraneous because x= y^2 = (-1.978719)^2 does not work. That leaves 2 sets of complex conjugates which would also be extraneous since I'm assuming we are looking for real solutions. I think the best way to approach this is first guess x=4. Then note that the graphs f(x)=x^3 and g(x)=sqrt(x)+62 intersect only in one place, so that x=4 is the only solution.

[@MichaelRothwell1]

Factoring y⁶-y-62 as you did is really helpful as we are just looking for non-negative roots and the second factor clearly has none.

[@alnitaka]

This problem is so hexic 9:52 it makes me feel depressed. Only one real root in radicals.

[@SyberMath]

🤩

[@SG49478]

Since you mentioned math competitions: Usually in the competitions I know are not allowing computers with Wolfram Alpha. And usually in these competitions they require proof, that you found all real solutions. That means your second method is elegant to find one solution, but is missing the proof, that the remaining part after your final factorization does indeed have no further real solutions. That would lead to deduction of at least half of the points for the problem. Here is how I approached it. I used mainly method 3, so substitution. y:=sqrt(x), we can also conclude that y>=0 because square roots are always non negative in the real world. With that we get y^6=y+62. y^6-y-62=0. With Vieta's formula we now just have to test 6 integers with 1;2;31 and -1;-2 and -31. Which works pretty quick since it is pretty obvious that 1;-1,31 and -31 don't work. We find y=2 is the only integer solution. Instead of using polynomial division or factoring by grouping we can also move forward using calculus proving, that the equation after substitution does not have any further solution for y>=0. If f(y)=y^6-y-62 => f'(y)=6y^5-1 and f"(y)=30y^4. 6y^5-1=0 has only one real solution which is 5throot(1/6). 30y^4 is always greater or equal to 0. With equaling 0 only for y=0. That means y=5throot(1/6) is a local minimum and the only extreme point. f(5throot(1/6)0. From that we can see: f has exactly one minimum at 5throot(1/6), no other extreme points and the value at the extreme point is negative. With f(y) becoming positive left from 5throot(1/6) and right from there it must intersect the x-axis exactly twice. With f(0) being still negative the intersection point on the left must be for a negative y value, therefor this is not a solution for our original equation. Therefor y^6-y-62=0 has only one real solution for non negative y. Therefor x=y^2=4 is the only solution of our original equation.

[@MichaelRothwell1]

Please see my comment for a similar solution but without calculus.

[@SG49478]

@@MichaelRothwell1 Your approach is another method which yields a complete solution with proof that there is no other. Well done!

[@pukulu]

x = 4 by inspection. Other roots are also possible.

[@emilywatt5126]

X=4 by instruction, then x3 is increasing function, and sqrt x +62 is also increasing with less speed and opposite curvature for all x>0. Therefore they intersect only in 1 point , and 4 is the only real solution

[@user-yr9zu3jm8f]

Không chỉ dạy toán mà thầy còn là người truyền cảm hứng và còn là người nâng tầm toán học thành một bộ môn nghệ thuật bất hủ

[@prollysine]

x^6-124x^3-x+3844=0 , by faktoring , (x-4)(x^5+4x^4+16x^3-60x^2-240x-961)=0 , x-4=0 , x=4 , OK , test , 4^3=64 , 4+62=64 ,

[@minecraftexplorer56]

It is just possible to guess. The number should any number's quadrate and cube.

[@MichaelRothwell1]

Using Method 3: To solve x³=√x+62 x=4 is a solution by inspection. Let y=√x (so y≥0) and we get y⁶=y+62 or y⁶-y-62=0, where y≥0. Let f(y)=y⁶-y-62, and we are looking for zeros for y≥0. For 0≤y≤1 we have y⁶≤1, -y≤0, so f(y)≤1+0-62=-611, f(y)=y⁶-y-62=y(y⁵-1)-62 y and y⁵-1 are both strictly increasing and >0, hence their product and f itself are both strictly increasing, so there is at most one solution. Hence y=2 is the only zero of f for y≥0 and x=4 is the only real solution to x³=√x+62.

[@ulrichwollandt3261]

4, just looking

[@Rbmukthegreat]

Great!

[@SyberMath]

Thank you!

[@2012tulio]

X=4

[@nasrullahhusnan2289]

By inspection x=4

[@user-fe5xs6yg6e]

Too easy to see the solution. 62 helps. It is almost 64. it lacks 2 and it is sqrt(x), so x may be 4. Check 4^3=64 - it works!. Now imagine graph of x^3 - sqrt(x) for x>0 and understand that the only real solution is possible.

[@rakenzarnsworld2]

x = 4

[@fahrenheit2101]

I swear you're making wayyyyyyy too many guess and check videos.

[@broytingaravsol]

made it

[@SidneiMV]

√x = u => x = u² u⁶ = u + 62 u⁶ - u - 62 = 0 u⁶ - 64 - (u - 2) = 0 (u² - 4)(u⁴ + 4u² + 16) - (u - 2) = 0 (u - 2)(u + 2)(u⁴ + 4u² + 16) - (u - 2) = 0 (u - 2)[(u + 2)(u⁴ + 4u² + 16) - 1] = 0 u - 2 = 0 => u = 2 => *x = 4* (u + 2)(u⁴ + 4u² + 16) - 1 = 0

[@SidneiMV]

(u + 2)(u⁴ + 4u² + 16) - 1 = 0 (u + 2)[(u² + 2)² + 12] - 1 = 0 .....