A Nice Olympiad Algebra Problem | X=? & Y= ? 

10:37 it should be m = -7, but luckily it didn't affect the square of m at 11:55 😊

Ah, I see someone else has also noted the Tutor's small error. M does not equal +7 but in fact equals -7. Still an excellent video! If this was a student's work very few marks should have been taken off, unless of course the error had made the rest of the working very much easier! I was told this at college many moons ago.

M= -7 Une erreur d'inattention qui peut tout fausser.

2X2 -4X-9=0

M=_7

m=-7

3 и 2

x=2; y=3;

m+7 =0, so m = -7, not 7

m=-7

Очень плохо. Допущены две ошибки. Такое впечатление, что стример зазубрил текст, не понимая смысла.

M+7=0. M= -7

Amazing that this can be done. However, I wish I could watch someone teach me how to do this type of math instead of just watching someone do the math. I can't learn it by watching the video. Nothing against the person doing this math. I just wish I could find a teacher that can teach what goes on in a person's head as you put a plan together and solve this type of math problem. Great job!

You must write equal sign on the fraction line level, not on the numerator level. This together with other mistakes mentioned here and idiotic thumbs in response to those mentions makes video ugly.

x³ + y³ = 35 { E1 } x² + y² = 13 { E2 } Note: x & y are interchangeable. Use z = x, y as roots of a quadratic equation: (z - x)*(z - y) = 0 z² - (x + y)*z + x*y = 0 { Note sum and product } z = ((x + y) ± √((x + y)² - 4*x*y))/2 { E3 } = ((x + y) ± √(x² + 2*x*y + y² - 4*x*y))/2 = ((x + y) ± √(x² - 2*x*y + y²))/2 = ((x + y) ± √(x - y)²)/2 = ((x + y) ± (x - y))/2 z = x, y { For "±", the "+" yields x, the "-" yields y. } Let s = (x + y) and p = (x*y) in E3: z = (s ± √(s² - 4*p))/2 { E3' } Let z⁺ = (s + √(s² - 4*p))/2 Let z⁻ = (s - √(s² - 4*p))/2 Thus, (x, y) = (z⁺, z⁻) Since x & y are interchangeable, this means: (x, y) = (z⁻, z⁺), (z⁺, z⁻) { E3" } Use E1 in the expansion of the cube of the sum: (x + y)³ = x³ + 3*x²*y + 3*x*y² + y³ (x + y)³ = (x³ + y³) + 3*x*y*(x + y) s³ = 35 + 3*p*s { E1' } Use E2 in the expansion of the square of the sum: (x + y)² = x² + 2*x*y + y² (x + y)² = (x² + y²) + 2*x*y s² = 13 + 2*p p = (s² - 13)/2 { E2' } Use E2' in E1' to eliminate p, and find s: s³ = 35 + 3*((s² - 13)/2)*s s³ = 35 + 3*s³/2 - 39*s/2 0 = 35 + s³/2 - 39*s/2 s³ - 39*s + 70 = 0 s³ - 4*s - 35*s + 70 = 0 s*(s - 2)*(s + 2) - 35*(s - 2) = 0 (s - 2)*(s² + 2*s - 35) = 0 s = 2, (-2 ± √144)/2 = 2, -1 ± 6 = 2, 5, -7 Use E2' in E3', and substitute s: z = (s ± √(s² - 4*(s² - 13)/2))/2 = (s ± √(s² - 2*(s² - 13)))/2 = (s ± √(26 - s²))/2 = (2 ± √22)/2, (5 ± 1)/2, (-7 ± i*√23)/2 From E3", we list the solutions: (x, y) = (z⁻, z⁺), (z⁺, z⁻) = (1 - √22/2, 1 + √22/2), (1 + √22/2, 1 - √22/2), (2, 3), (3, 2), (-7/2 - i*√23/2, -7/2 + i*√23/2), (-7/2 + i*√23/2, -7/2 - i*√23/2) Alternative solution: x³ + y³ = 35 { Note: 3³ + 2³ } x² + y² = 13 { Note: 3² + 2² } x & y are interchangeable. y⁶ = y⁶ (y²)³ = (y³)² (13 - x²)³ = (35 - x³)² 13³ - 3*13²*x² + 3*13*x⁴ - x⁶ = 1225 - 70*x³ + x⁶ 2*x⁶ - 39*x⁴ - 70*x³ + 507*x² - 972 = 0 (x² - 5*x + 6)*(2*x² - 4*x - 9)*(x² + 7*x + 18) = 0 x = (5 ± 1)/2, 1 ± √22/2, (-7 ± i*√23)/2 y = (5 ∓ 1)/2, 1 ∓ √22/2, (-7 ∓ i*√23)/2 (x, y) = (2, 3), (3, 2), (1 - √22/2, 1 + √22/2), (1 + √22/2, 1 - √22/2), ((-7 - i*√23)/2, (-7 + i*√23)/2), ((-7 + i*√23)/2, (-7 - i*√23)/2) Full factorization was done via synthetic division. x = 2, 3 , so a factor is (x - 2)*(x - 3) = x² - 5*x + 6 . 1 -5 6 * 2 10 -1 -135 -162 2 0 -39 -70 507 0 -972 2 -10 12 0 10 -51 -70 507 0 -972 0 10 -50 60 0 0 -1 -130 507 0 -972 0 0 -1 5 -6 0 0 0 -135 513 0 -972 0 0 0 -135 675 -810 0 0 0 0 -162 810 -972 0 0 0 0 -162 810 -972 Factor 2: 1*2 . Factor -162: -2*3⁴ = -9*18 . 2 -4 -9 * 1 7 18 2 10 -1 -135 -162 2 -4 -9 0 14 8 -135 -162 0 14 -28 -63 0 0 36 -72 -162 0 0 36 -72 -162 Other factorizations: (2*x³ - 10*x² + 3*x + 27)*(x³ + 5*x² + 4*x - 36) = 0 (2*x³ - 8*x² - x + 18)*(x³ + 4*x² - 3*x - 54) = 0 Commentary on video: 07:45 (m² - 4) isn't a perfect square 09:23 you can complete the square to solve faster: m² + 2* m + 1 = (m + 1)² = 36 m = -1 ± 6 = 5, -7 10:37 you wrote 7 instead of -7 13:30 m is the sum, and n is the product. They're the coefficients of a quadratic equation: z² - m*z + n = 0 This is readily solved via the quadratic formula: z = (m ± √(m² - 4*n)/2 You can put in the m & n values directly and simplify to get the roots immediately. 15:09 you wrote x instead of x² You previously had mispronounced it "eksy", and later mispronounce discriminant as "de-criminint".

"perfect square" is formula for expansion of (a+-b)**2 into a**2+-2*a*b+b**2 and back formula for a**2-b**2 = (a+b)*(a-b) is called "difference of squares"

10:38 wrong, that should equal to -7.

M^3-3MN=35， M^2-2N=13 ➡️ 2M^3-6MN=70， 3M^3-6MN=39M，➡️M^3-39M+70=0

correction m2=-7 and 2x^2-4x-9 =0 not 2x-4x-9=0

NEW WAY

M = -7 !!!!!

m=-7

M+7=0. M=-7

The methodology consists of decomposing the the powers in terms that could be factored and identify them with the given equations the additional trick is to create the new variables for easy manipulation

15:10 should be 2x^2

Сделать замену ху=а, х+у=с, тогда уравнение 1 имеет вид с*(13-а)=35 Уравнение 2 с^2-2а=14 Тогда с*(40-с^2)=70

In the second possible solution (m) will be equal to (-7) not (7)

And also it should be 2x^2 at 15:08

Think two less than 13 such that sum of their square equals 13.2and 3 sre such twonumbers. The sum of the cubes of these numbers equals 35 Hence x = 2or3 while y= 3 or 2 are the solution of these two equations. P.k.Malhotra

X=3, y=2 0:05

いかにも簡単に見えて難しいですね。xが３でyが２でしょうね。

This level of mathematics is secondary or university

X=3,.Y=2.

m=-7

M=😮

X=2,Y =3

2 and 3. In three seconds with trial and error