Problem indicated is different from what is worked out
@user-bk2bi4rv4h3 күн бұрын
問題が間違っています。 前が+、後が-
@tarunjain15373 күн бұрын
Is he teaching maths or verses from quran?
@echandler3 күн бұрын
While solving a quadratic for a constant is novel, it obscures the heart of the problem here. The flavor of this is that of a function and its inverse. Set a new variable y to LHS and RHS, then subtract the two resulting equations. Lastly factor and solve from there. (17+x)(17-x) = √(17^2-x) = y 17^2-x^2 = y ⇒ x^2 = 17^2- y #1 square the second set y^2 = 17^2 - x subtract and factor x^2 - y^2 = x-y (x-y)(x+y)-(x-y) = 0 (x-y)(x+y-1) = 0 Hence y=x *or* y=1-x Substitute in #1 and solve the resulting quadratics. This pattern can be applied to other similar contest problems.
@guyhoghton3993 күн бұрын
Interesting technique, thank you. I think the following approach is simpler for this case however. Let _y = √(289 - x)_ ∴ _y² = 289 - x_ ⇒ _289 - y² = x_ ... ① The equation is: ∴ _(17 + x)(17 - x) = 289 - x² = y_ ... ② Subtract ② from ①: _x² - y² = x - y_ ⇒ _(x - y)(x + y - 1) = 0_ ⇒ _x = y or x = 1 - y_ (i) *_x = y_* ⇒ _x² = y² = 289 - x_ ⇒ _x² + x - 289 = 0_ ⇒ *_x = ½( -1 ± √[1 - (4)(-289)] ) = ½( -1 ± √1157 )_* (ii) *_x = 1 - y_* ⇒ _x = 1 - (289 - x²)_ (from ②) ⇒ _x² - x - 288 = 0_ ⇒ *_x = ½( 1 ± √[1 - (4)(-288)] ) = ½( 1 ± √1153 )_* Then filter (i) and (ii) by _-17 < x < 17_
Expanding with binomial theorem, canceling highest and lowest degree terms, and dividing through by the spurious factor of 7x (as x ≠ 0) leaves: 0 = x⁵ + 3x⁴ + 5x³ + 5x² + 3x + x. Then, noting the symmetry between paired odd and even degree terms, 0 = (x + 1) ⋅ (x⁴ + 2x³ + 3x² + 2x + 1). Now again recognizing the symmetry, and so temporarily factoring out x², gives: 0 = (x + 1) ⋅ x² ⋅ ( x² + 2x + 3 + 2(1/x) + (1/x)² ) and on regrouping 0 = (x + 1) ⋅ x² ⋅ ( [x² + (1/x)² + 2] - 2 + 2⋅[x + (1/x)] + 3 ) = (x + 1) ⋅ x² ⋅ ( [x + (1/x)]² + 2⋅[x + (1/x)] + 1 ) = (x + 1) ⋅ x² ⋅ ( [x + (1/x)] + 1 )² = (x + 1) ⋅ ( x² + x + 1 )² from which all solutions are readily obtained. One of the most important concepts you could be teaching, through demonstration, is how to present work in a clear, concise, and structured easy-to-read fashion. You tag your problems as Olympiad problems, but then side track into elementary and intermediate algebra details that the target audience is already expert in - or is working to become expert in. Help them out with that.
@key_board_x8 күн бұрын
1 + (1/x⁷) = [1 + (1/x)]⁷ 1 + (1/x)⁷ = [1 + (1/x)]⁷ → let: a = 1/x → where: x ≠ 0 1 + a⁷ = (1 + a)⁷ 1 + a⁷ = (1 + a)².(1 + a)².(1 + a)².(1 + a) 1 + a⁷ = (1 + 2a + a²).(1 + 2a + a²).(1 + 2a + a²).(1 + a) 1 + a⁷ = (1 + 2a + a² + 2a + 4a² + 2a³ + a² + 2a³ + a⁴).(1 + a + 2a + 2a² + a² + a³) 1 + a⁷ = (1 + 4a + 6a² + 4a³ + a⁴).(1 + 3a + 3a² + a³) 1 + a⁷ = 1 + 3a + 3a² + a³ + 4a + 12a² + 12a³ + 4a⁴ + 6a² + 18a³ + 18a⁴ + 6a⁵ + 4a³ + 12a⁴ + 12a⁵ + 4a⁶ + a⁴ + 3a⁵ + 3a⁶ + a⁷ 0 = 7a + 21a² + 35a³ + 35a⁴ + 21a⁵ + 7a⁶ 0 = 7.(a + 3a² + 5a³ + 5a⁴ + 3a⁵ + a⁶) a + 3a² + 5a³ + 5a⁴ + 3a⁵ + a⁶ = 0 a.(a⁵ + 3a⁴ + 5a³ + 5a² + 3a + 1) = 0 First case: a = 0 Second case: (a⁵ + 3a⁴ + 5a³ + 5a² + 3a + 1) = 0 → you can see that (- 1) is an obvious root, so you can factorize (a + 1) (a + 1).(a⁴ + αa³ + βa² + γa + 1) = 0 → you expand a⁵ + αa⁴ + βa³ + γa² + a + a⁴ + αa³ + βa² + γa + 1 = 0 → you group a⁵ + a⁴.(α + 1) + a³.(β + α) + a².(γ + β) + a.(1 + γ) + 1 = 0 → you compare to: (a⁵ + 3a⁴ + 5a³ + 5a² + 3a + 1) (α + 1) = 3 → α = 2 (β + α) = 5 → β = 3 (γ + β) = 5 → γ = 2 (1 + γ) = 3 → γ = 2 → of course, above The equation becomes: (a + 1).(a⁴ + 2a³ + 3a² + 2a + 1) = 0 (a + 1).(a⁴ + a³ + a³ + a² + a² + a² + a + a + 1) = 0 (a + 1).(a⁴ + a³ + a² + a³ + a² + a + a² + a + 1) = 0 (a + 1).[(a⁴ + a³ + a²) + (a³ + a² + a) + (a² + a + 1)] = 0 (a + 1).[a².(a² + a + 1) + a.(a² + a + 1) + (a² + a + 1)] = 0 (a + 1).(a² + a + 1).(a² + a + 1) = 0 (a + 1).(a² + a + 1)² = 0 First case: a = - 1 Second case: (a² + a + 1) = 0 Δ = (1)² - (4 * 1) = - 3 = 3i² a = - 1 ± i√3 Resume the cases: a = 0 a = - 1 a = - 1 + i√3 a = - 1 - i√3 Recall: a = 1/x → x = 1/a → where a ≠ 0 First: a = 0 → no possible because the condition Second: a = - 1 → x = 1/a = - 1 Third: a = - 1 + i√3 x = 1/(- 1 + i√3) x = (- 1 - i√3)/[(- 1 + i√3).(- 1 - i√3)] x = (- 1 - i√3)/[1 - 3i²] x = (- 1 - i√3)/4 Fourth: a = - 1 - i√3 x = 1/(- 1 - i√3) x = (- 1 + i√3)/[(- 1 - i√3).(- 1 + i√3)] x = (- 1 + i√3)/[1 - 3i²] x = (- 1 + i√3)/4 Solution = { - 1 ; (- 1 - i√3)/4 ; = (- 1 + i√3)/4 }
@pietergeerkens63248 күн бұрын
For the binomial expansion, just use Pascal's Triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 6 1 instead off all that nit-picky multiplication.
1/x = u => x = 1/u 1 + u⁷ = (1 + u)⁷ 7(u⁶ + u) + 21(u⁵ + u²) + 35(u⁴ + u³) = 0 (u⁵ + 1) + 3u(u³ + 1) + 5u²(u + 1) = 0 (u + 1)[(u⁴ - u³ + u² - u + 1) + 3u(u² - u + 1) + 5u²] = 0 (u + 1)[(u⁴ - u³ + u² - u + 1) + 3u(u² - u + 1) + 5u²] = 0 (u + 1)(u⁴ + 2u³ + 3u² + 2u + 1) = 0 u = -1 => *x = -1* u⁴ + 2u³ + 3u² + 2u + 1 = 0 u² + 1/u² + 2(u + 1/u) + 3 = 0 (u + 1/u)² + 2(u + 1/u) + 1 = 0 u + 1/u = w w² + 2w + 1 = 0 (w + 1)² = 0 w = - 1 => u + 1/u = -1 => x + 1/x = -1 x² + x + 1 = 0 *x = (-1 ± i√3)/2*
@pietergeerkens63249 күн бұрын
I believe the following is easier to follow, and with much smaller magnitude constants makes the arithmetic and algebra simpler. This is a valuable consideration under exam or contest conditions, as increasing confidence that one's algebra is correct on the first pass. Dividing through by 4 and letting u = √x / 2¹⁰ yields: ⁵√[ ½ + u ] + ⁵√[ ½ - u ] = 1 Then defining a = ⁵√[ ½ + u ] b = ⁵√[ ½ - u ] v = ab = ⁵√[ ¼ - u² ] we obtain a + b = 1 ab = v and a² + b² = (a+b)² - 2ab = 1 - 2v a³ + b³ = (a+b)³ - 3ab(a+b) = 1 - 3v and finally that 1 = a⁵ + b⁵ = (a² + b²)(a³ + b³) - (ab)²(a+b) = 1 - 5v + 6v² - v² or 5v² - 5v = 5⋅v⋅(v-1) 0. Now v = 1 can be rejected as requiring u² < 0, and v = 0 gives u² = ¼ and finally x = (2¹⁰)²⋅u² = 2¹⁸.
@SidneiMV9 күн бұрын
x + 9 = u => x = u - 9 (u - 8)³ + (u - 4)³ + u³ + (u + 4)³ + (u + 8)³ = 10³ u³ + 2(u³ + 3u8²) + 2(u³ + 3u4²) = 10³ 5u³ + 480u - 1000 = 0 u³ + 96u - 200 = 0 u³ - 8 + 96(u - 2) = 0 (u - 2)(u² + 2u + 100) = 0 u - 2 = 0 => u = 2 => *x = -7* -6³ + -2³ + 2³ + 6³ + 10³ = 10³
@pietergeerkens63249 күн бұрын
Nice problem; unimpressive solution. Always look for symmetries on complex problems! This one took only about 20 seconds to read, solve, and prove with just mental math. Here: (x+1)^3 + (x+13)^3 = 0 when x = -7 as (-6)^3 + (6)^3 = 0 (x+5)^3 + (x+ 9)^3 = 0 when x = -7 as (-2)^3 + (2)^3 = 0 Unsurprisingly, this solves the problem as 10^3 = 10^3. To succeed on Olympiad level problems, candidates must be trained to look for, and recognize, this level of analysis. The techniques you use here should be demonstrated on problems that actually require them.
Ar fi bine ca enunțul sa fie corect 160 x nu 160 Ok,
@freddyalvaradamaranon30414 күн бұрын
Interesante ejercicio de ecuacion cuadratica, gracias por compartir 😊😊. Buena y didáctica explicación, utilizando propiedades de radicales y ecuaciones cuadraticas.😊😊.