Відео

I did not solve this problem but I guessed early on the solution lies between 0 & 0.4.

Real root : x=3 ,complex root solution..x=3-+.. Thanks for sharing

X=3

Awesome. Is there a general method to denest sqrt(cbrt(a)-cbrt(b)) for example sqrt(cbrt(5)-cbrt(2))?

real root: x=3, complex roots x=3±i√6

3

Thanks....it was a wonderful explanation

Wonderful and beautiful!

Για μενα αυτη ηταν μια δυσκολη ασκηση. 10 χρονια να παιδευομουνα δεν θα μπορουσα να σκεφτω τη λυση σας.

Excellent

Thanks for sharing.....it was a wonderful explanation....x=4

Pass! Must be a Gén0cide math if it is from that country.

(8^t, 4^t, 2^t) > 0. If integers: 1 + 1 + 4 = 6. x = -2 does the trick. May be other roots.

First of all, the claim in the video at 0:40 that the graph of y = 1/√x is a hyperbola is _wrong_ because this is actually one branch of the graph of xy² = 1 which is a _cubic_ curve, not a quadratic curve. The graph of y = 1/x or xy = 1 on the other hand _is_ a hyperbola but like all conic sections this is a _quadratic_ curve. The equation to solve in ℝ is x² + 5x + 5 = 1/√x Substituting x = a² and therefore √x = a assuming a > 0 and then multiplying both sides by a we have (1) a⁵ + 5a³ + 5a = 1 This equation is a special type of quintic equation which can be solved by a hyperbolic substitution. If we substitute (2) a = 2·sinh(t) we have (3) 32·sinh⁵(t) + 40·sinh³(t) + 10·sinh(t) = 1 Dividing both sides by 2 this gives (4) 16·sinh⁵(t) + 20·sinh³(t) + 5·sinh(t) = ½ and using the identity (5) sinh(5t) = 16·sinh⁵(t) + 20·sinh³(t) + 5·sinh(t) we can write (4) as (6) sinh(5t) = ½ Since the hyperbolic sine is a periodic function with period 2πi this gives (7) t = ⅕·arsinh(½) + k·⅖πi, k ∈ ℤ Substituting (7) in (2) the roots of the equation (1) are (8) a = 2·sinh(⅕·arsinh(½) + k·⅖πi), k = 0..4 The equation has five roots, four of which are complex. Setting k = 0 gives the only real root, which is (9) a = 2·sinh(⅕·arsinh(½)) To write this root in algebraic form, we may note that 2·sinh(t) = exp(t) − exp(−t) and arsinh(u) = ln(u + √(u² + 1)). So, we have arsinh(½) = ln(½ + ½√5) and since (½√5 + ½)(½√5 − ½) = 1 we also have −arsinh(½) = −ln(½ + ½√5) = ln(½√5 − ½). Therefore, the real root is a = 2·sinh(⅕·arsinh(½)) = exp(⅕·arsinh(½)) − exp(−⅕·arsinh(½)) = exp(⅕·ln(½ + ½√5)) − exp(⅕·ln(½√5 − ½)) = ⁵√(½ + ½√5) − ⁵√(½√5 − ½) = ⁵√(½ + ½√5) + ⁵√(½ − ½√5) and since x = a² this gives x = (⁵√(½ + ½√5) + ⁵√(½ − ½√5))² Noting that (½ + ½√5)² = ³⁄₂ + ¹⁄₂√5, (½ − ½√5)² = ³⁄₂ − ¹⁄₂√5 and (½ + ½√5)(½ − ½√5) = −1 the exact solution can be written as x = −2 + ⁵√(³⁄₂ + ¹⁄₂√5) + ⁵√(³⁄₂ − ¹⁄₂√5) The numerical value of this solution accurate to seven decimal places is 0.0371649.

A wonderful introduction.👌.....x=-2 is a real solution

Too nice explanation...E= 1873...

Like@ashokdubey

Let a = 2^(2x+4). Then, the given equation is written as a^3+2a^2=3. This has a=1 as the only real solution. So, 2x+4=0 > x = -2.

Manipulating and letting t = 4^(x +2), the given becomes t^3 +t^3 +4t^2 =6; 2t^3 +4t^2 = 6; t^3 +2t^2 -3 =0. By RRT and SDM, (t -1)(t^2 +3t +3) =0, that is t -1 =0; t = 1 and t^2 +3t +3 =0; cmplx sols. Thus, t = 4^(x +2) = 1 = 4^0; x +2 = 0; x = -2

Let ln x = a, lny=b and lnz=c. Then ln2+a+b = ab, b+c=bc, ln2+c+a=ca. Therefore (a-1)(b-c)=0. But a= 1 gives ln2+1=0, which is not possible. So, b=c. Thus, 2b=b^2 > b=0,2. If b=c=0, y=z=1 and a=-ln2 > x=1/2. So, (x,y,z) = (1/2,1,1) If b=c=2, a=ln2 + 2. Thus, x=2e^2, y=e^2, z=e^2 , e^2(2,1,1). If the log is to the base 10, we have 100(2,1,1).

x=-2

Thanks for sharing....x=3

Thank you Sir...x=0,037 only one real solution

(x,y,z)=(1/2,1,1),(200,100,100)

Thanks

Nice solution. Some remark: numerator and denominator should be multiplied by 2 as we add a^6 to a^6 and subsequent similar terms. Of course this 2 will be cancelled but we should include it.

x=0.037

Let x-3=t. The equation becomes (2^6-1)t^6 +(60)(15)t^4 +(48)(15)t^2 = 0. So, either t=0 > x=3 or 7t^4 + 100t^2 +80 = 0. This does not have positive solutions for t^2 (x cannot be real). So, x=3.

X=3

Sub x = y + 3. Converts then to ((y+2)^6 + (y-2)^6) / ((y+1)^6 + (y-1)^6) = 2^6. At y = 0, (2 * (2^6)) / (2 * 1^6) = 2^6. check. Then X = 0 + 3 = 3. As the shape of the relationship is maximum at y = 0, no other real roots exist.

X= 3

Very good

Multiplying 4 and adding 5 in the process was tricky but wonderful, sir ^.^

I get the same solutions by squaring both sides and solving (after rearranging the terms) the resultant polynomial x^4+8x^3+9x^2+18x+10=0. Clearly (x-1) is a factor so after division we're left with x^3+9x^2+8x+10=0. (x+1) is a factor so after further division we have x^2+8x+10=0 with solutions x=-4+-SQRT(6). Remaining solutions are, of course, x=1, x=-1

A wonderful introduction and clearly explaining...thanks....x=y=z= 1^2/8 ....

=1873

1/2

another approach let a = x^3 +1, y^4 + 1 = a(a+1) => a^2 + a - (y^4 +1) = 0 discriminant D = 4y^4 + 5 = k^2 for some positive integer k 4y^4 - k^2 = (2y^2 + k) * (2y^2 - k) = -5 1st term is positive, only (5, -1) and (1,-5) are possible cases. case1 (2y^2 + k) = 5, (2y^2 - k) = -1 => y = ±1, k = 3 => a = (-1 ± k)/2 = 1 or -2 a = x^3 + 1 = 1 or -2 => x = 0 is only integer solution. (x, y) = (0, ±1) case2 (2y^2 + k) = 1, (2y^2 - k) = -5 => y^2 = -1 => no integer(real) solution.

My solution, before being able to watch the video: The equation to solve is (x + 2)² = √(15x² + 40x + 26) Square both sides and we get (x + 2)⁴ = 15x² + 40x + 26 x⁴ + 8x³ + 24x² + 32x + 16 = 15x² + 40x + 26 x⁴ + 8x³ + 9x² − 8x − 10 = 0 x⁴ + 8x(x² − 1) + 9x² − 10 = 0 x²(x² − 1) + 8x(x² − 1) + 10x² − 10 = 0 x²(x² − 1) + 8x(x² − 1) + 10(x² − 1) = 0 (x² − 1)(x² + 8x + 10) = 0 (x − 1)(x + 1)((x + 4)² − 6) = 0 x − 1 = 0 ⋁ x + 1 = 0 ⋁ (x + 4)² − 6 = 0 x = 1 ⋁ x = −1 ⋁ x = −4 + √6 ⋁ x = −4 − √6 So we have four real solutions, but now we still need to verify these solutions because the radicand 15x² + 40x + 26 at the right hand side of the original equation must not be negative. It is easy to see that this condition is satisfied for x = 1 and x = −1, but what about the irrational solutions x = −4 + √6 and x = −4 − √6 which are both negative? Note that you are _not allowed to use a calculator_ when solving problems that appeared on math contests. We can proceed as follows. Note that x = −4 + √6 and x = −4 − √6 are both solutions of the quadratic x² + 8x + 10 = 0 so for each of these values of x we have x² = −8x − 10 and therefore 15x² + 40x + 26 = 15(−8x − 10) + 40x + 26 = −80x − 124 So, for x = −4 − √6 we have 15x² + 40x + 26 = −80x − 124 = −80(−4 − √6) − 124 = 196 + 80√6 which is evidently positive. But for x = −4 + √6 we have 15x² + 40x + 26 = −80x − 124 = −80(−4 + √6) − 124 = 196 − 80√6 and it is not immediately obvious whether this is positive or negative. So how do we decide this _without_ using any calculating device? Well, we have (196 + 80√6)(196 − 80√6) = 196² − (80√6)² = (200 − 4)² − 6400·6 = 40000 − 1600 + 16 − 36000 − 2400 = 40000 − 36000 − 4000 + 16 = 16 So, since 196 + 80√6 is positive and since the product of 196 + 80√6 and 196 − 80√6 is positive, this means that 196 − 80√6 itself must also be positive. So, for x = −4 + √6 the radicand 15x² + 40x + 26 = 196 − 80√6 is indeed positive. Therefore, all four solutions x = 1 ⋁ x = −1 ⋁ x = −4 + √6 ⋁ x = −4 − √6 are accepted as solutions of the equation (x + 2)² = √(15x² + 40x + 26).