Video

my bro was flabbergasted with the second method. quote: "NO ONE REMEMBERS THE CUBIC BINOMIAL!!"

Let f(x) be the left-hand side of this equation. It is easy to show that f(x) is concave up everywhere and has a minimal point at x=0 and f(0)=3.

If you let a/b=u, then the equation becomes (u^3+3u)/(3u^2+1)=63/62, or u=triscoth(63/62). The only problem is hunting a rational root. Just solve 62(u^3+3u)=63(3u^2+1)

*What about complex solutions?*

😊🎉👍👍👍🎉😊

Hhh

once you have (a+b)/(a-b) = 5 = 5/1, you can use componendo-dividendo again or *reverse* componendo-dividendo. (5+1)/2 = 3, (5-1)/2 = 2 thus (a+b)/(a-b) = (3+2)/(3-2) hence a/b = 3/2.

😮

Solve A Beautiful Equation: xˣ = 2¹⸍ˣ; x = ? x ≠ 0; (xˣ)ˣ = (2¹⸍ˣ)ˣ, x^x² = 2, (x^x²)² = 2², (x²)^x² = 2², x² = 2; x = ± √2 Answer check: 2¹⸍ˣ = 2^(± 1/√2) = [(√2)²]^(± 1/√2) = (√2)^(2/√2) = (√2)^(√2) = xˣ; Confirmed Final answer: x = √2 or x = - √2

3/2

62 b³ ((a/b)³ + 3 (a/b)) = 63 a³ ((b/a)³ + 3 (b/a)) and now with c = a/b we get: 62c³ - 189c² + 186x -63 = 0 (there is a typo in the video: -129 instead of -189) (2c - 3) (31c² - 48c + 21) = 0 with one real solution c= 3/2 and two complex ones.

The x/y = z/w => (x+y)/(x-y) = (z+w)/(z-w) works because x/y = z/w x = zy/w x+y = y+(zy)/w (x+y)/(x-y) = ((zy/w)+y)/(x-y) Knowing that y = xw/z, we can replace (x+y)/(x-y) = xw(z+w)/(zx(1-w/z)) (x+y)/(x-y) = (z+w)/(z-w)

Typo at 7:40. The coefficient of u^2 is -189.

😊😊😊👍👍👍

x² - 18x = 17√x => (x² - x) - (17x + 17√x) = 0 a² - b² = (a-b)(a+b) If a = x, b = √x then (x² - x) = (x - √x)(x + √x) and (x - √x)(x + √x) - 17(x + √x) = 0 or (x + √x)*(x - √x - 17) = 0 Therefore 1) x + √x = 0 => x = 0 and x - √x = 0 2) x - √x = 17

x=1 and y=0 also works

x² - 18x = 17√x find E = x - √x √x⁴ - 18√x² = 17√x √x(√x³ - 18√x - 17) = 0 √x = 0 => x = 0 => *E = 0* √x³ - 18√x - 17 = 0 √x³ + 1 - 18√x - 18 = 0 (√x + 1)(√x² - √x + 1) - 18(√x + 1) = 0 √x² - √x - 17 = 0 => *E = x - √x = 17*

Let A=x-√x; A*(x+√x) = x^2 - x; Eqn: x^2 - x = 17(x+√x); A*(x+√x) =17( 17(x+√x); A = 17

Another way would be just calling y = sqrt(x). This gives us the equation y⁴-18y²-17y=0. By inspection, we quickly notice that y=-1 is a possibility. So, by Briot-Ruffini, we have y³-y²-17y=0, or y(y²-y-17)=0, and therefore, y=0 or y=(1±sqrt(69))/2. Returning to x, as we have y = sqrt(x), the only admissible values are those who are ≥0. So sqrt(x) can be 0 (and then, x=0, and therefore, x-sqrt(x)=0) or (1+sqrt(69))/2 (and then, x=(35+sqrt(69))/2, and therefore, x-sqrt(x)=17).

Method 2 slightly quicker if you factor out sqrt(x)+1 sooner.

Zero is asnwer too.

The first method gets easier if we use √x = y as a substition and so y⁴ - 18y² - 17y = 0. y=0 (and so x=0) is a solution we already know so we can divide by y: y³ - 18y - 17 = 0 = (y + 1) (y² - y - 17) y must be positive for real solutions so the only possible y value is (1 + √69) / 2 which can easily be squared to get x and so x - √x = y² - y = 17.

17

At 2:43: why?

X= 0, Y=0.

Proving that the correct answer is the golden ratio, phi, can be done quickly. Fact: phi^n = (Ln + Fn sqrt(5))/2, Ln and Fn are Lucas and Fibonacci numbers. Since L6 = 18, F6 = 8, phi^6 = 9 + 4sqrt(5), and we are done.

Nice

TKALSS 😁

😊😊😊👍👍👍

u = 2^(x^2) u + u^2 = u^3 1 + u = u^2 u = (1 + √5)/2 x^2 = ln[(1 + √5)/2] / ln(2) x = ±√{ln[(1 + √5)/2] / ln(2)}

x³+x²+4=(x²+ax+b)(x+c) =x³+(c+a)x²+(ac+b)x+bc c+a=1 (1) ac+b=0 (2) bc=4 (3) Multiply (2) by c: ac²+bc=0 Noting (3) ac²+4=0 ac=-4 (4) Multiply (1) by c²: c³+ac²=c² Noting (4) c³-4=c² c²(c-1)=4 =2²(2-1) c=2, b=2, a=-1 Thus 0=x³+x²+4 =(x²--x+2)(x+2) x=-2 or x²-x+2=0 x=½[1±sqrt(-7)] =½[1±isqrt(7)]

Due to small constant, by scrutinizing the equation it is clear that x=-2

For the 2nd method, to avoid invalid solution value w/o checking, note that x+y=(x-y)² --> x+y is positive It means that • both x and y can't be negative • both x and y are positive or • one of x and y is positive and the other one is negative but the absolute value of the negative one is less than the positive. To be clearer, let say y<0. Then |y|<x. Hope tham I'm right.

The unsaid fact: n(n+1)/2 is the nth triangular number. Also, any time 8n+1 (where n is any variable or expression representing an integer) turns up alone under a square root, you can be sure that triangular numbers are lurking somewhere. A bit like if n^2-n-1 shows up then there's usually a golden ratio afoot.

Nice problem and easy so solve. 😊 btw: at 7:04 you wrote root x² - 1 instead of root x² + 1

In #2 k and -k give (x,y) and (y,x)

took me a while of bashing my skull in to the wall with congruence, parity, perfect squares and divisibility, but then the solution hit me and i felt kind of dumb.. Set x-y=n, therefore x=y+n Replace in our original equation to get n+y+y=n^2 Cleaning up you get y = ½(n^2 - n) Do the same for y=x-n, x= ½(n^2 + n) The solution set to our original equation is (x, y) = [½(n^2 + n), ½(n^2 - n)] for every integer n

*x and y are both equal to 0*

If x + y = k^2 then x - y can be k or (-k). Hence the 2 solutions you got with the 1st method.

only for (0, 0)?

x ∈ { -√[ln[½(1+√5)]/ln 2], √[ln[½(1+√5)]/ln 2], -√{ln[½(√5-1)]+iπ(1+2k)}/ln 2}, √{ln[½(√5-1)]+iπ(1+2k)}/ln 2}, ( k ∈ ℤ ) }

substitute f(t) and solve: √x²+1 = (t+1)x > 0 x²+1 = (t+1)²x² ((t+1)²-1)x² =1 x² = 1/(t²+2t) so f(t)= 1/(t²+2t)

Let y=x² The given equation becomes (2^y)+(4^y)=8^y --> (2^y)+(2^y)²=(2^y)³ Divide 2^y≠0 then put all terms in one side: (2^y)²-(2^y)-1=0 is a equation for golden ratio ß, say 2^y = ß where ß=½(1+sqrt(5)] y=²log(ß) where the log is based 2 x=±sqrt[²log(ß)] The other root, -1/ß doesn't apply as 2^y can't be negative

Notice that a+1|2(a+1) so a+1|2a+7-(2a+2)=5 therefore a+1|5 which means a=-6,-2,0,4 so b=1,-3,7 and 3.

How I did it. Assume x>0, worry about negative x later. Multiply input of x by its conjugate, top and bottom. Change x to sqrt(x). This gives f(1/(sqrt(x^2 + x) + x)) = x. inverse of both sides: 1/(sqrt(x^2 + x) + x = f^-1(x) = y. SWAP x <-> y 1/(sqrt(y^2 + y) + y = f^-1(y) = x Solve for y = f(x). sqrt(y^2 + y) = 1/x - y y^2 + y = 1/x^2 - 2y/x + y^2 The y^2 cancel out! y = 1/(x^2(1+2/x) = 1/(x^2 + 2x) = f(x) Your way was easier

What are the answers in binary considering that you were using base 2?

easy, peasy!

Nice

2^x² = u (u > 0) u² - u - 1 = 0 u = (1 + √5)/2 2^x² = (1 + √5)/2 x²ln2 = ln( 1 + √5) - ln2 x² = (1/ln2)ln(1 + √5) - 1 x = ± √[ (1/ln2)ln(1 + √5) - 1 ]

Goal: find all functions f: ℝ→ℝ such that for all x∈ℝ* ; f([√(x^2+1) − x]/x) = x^2 One first notices that x = √(x^2) if x>0 and that x = −√(x^2) if x<0 . We substitute x with these expressions respectively and then notice that for all x∈ℝ ; t = x^2 ∈ ℝ>0 , which means the equation is equivalent to the following: For all t ∈ ℝ>0 ; f([√(t+1) − √t]/√t) = t and f([√(t+1) + √t]/−√t) = t. Let's solve the 1st part. Next, let u = [√(t+1) − √t]/√t ie. u√t = √(t+1) − √t ie. (u+1)√t = √(t+1) ie. (u+1)^2×t = t+1 and t+1≥0 (which is verified because t ∈ ℝ>0) ie. [(u+1)^2−1]×t = 1 ie. t = 1/[(u+1)^2−1] = 1/[u(u+2)]. For all t ∈ ℝ>0 ; t+1>t which implies √(t+1)>√t ⇒ √(t+1) −√t >0 and 1/√t >0 ⇒ u = [√(t+1) − √t]/√t ∈ℝ>0 so the 1st part becomes: For all u ∈ ℝ>0 ; f(u) = 1/[u(u+2)] Similarly, let v = [√(t+1) + √t]/−√t ie. v√t = −√(t+1) − √t ie. (v+1)√t = −√(t+1) ie. (v+1)^2×t = t+1 and t+1≥0 ... same work as above ... t = 1/[v(v+2)]. For all t ∈ ℝ>0 ; t+1>1 and t>0 which implies √(t+1)>1 and √t>0 ⇒ √(t+1) +√t >1 and 1/√t >0 ⇒ [√(t+1) + √t]/√t >0 ⇒ v = [√(t+1) + √t]/−√t ∈ℝ<0 so the 2nd part becomes: For all v ∈ ℝ<0 ; f(v) = 1/[v(v+2)] u and v are dummy variables we could call them anything as long as it is from their ranges. The final solution is any function f such that for all x∈ℝ\{0,−2} ; f(x) = 1/[x(x+2)] and f(0) = a and f(−2) = b where a , b ∈ℝ.