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Good, but too much redundant works! Once you've got your (eq.3) a+b=-1, then just substituting it into either of the given eqns (eq.1 or .2) yields the same results simply.
Agree with this. Another way to make it more efficient (though less than above, but you dont have to solve the quadratic) is once you have arrived at a+b=-1 and ab=-132, simply consider the possible factors of 132, and deduce the (obvious) solutions.
В России за такое извращение учителя будут ругаться на учеников)) метод подстановки, ну вырази а через b и всё...каждый раз на этом канале удивляюсь. Простейшие задачи называют олимпиадными и решаются максимально криво. С одной стороны, весело за таким наблюдать. С другой стороны, это очень печально..
Nice solution, but looks like it is from some ancient world, where no formula for quadratic equation roots is known yet ). It could have been done in a more straightforward way: after we got equation a+b+1=0 we get a= -1-b from here and then substitute a in Eq #2 with this. We get then b^2 +b+1=133. From here we get b^2 + b -132=0. And we just find roots using standard way, via discriminant.
Aquí tenemos un ejemplo de la diferencia entre eficacia y eficiencia. La resolución fué eficaz pues se llegó al resultado correcto; pero no fue eficiente pues no utilizó los conocimientos que le permitirían llegar con menos recursos y muchísimo más rápido.
a^2 - b = 133 = b^2 - a, so a^2 - b^2 = b - a, so (a+b)(a-b) = -1(a-b), so a+b = -1, since a NOT = b. So, a and b are the 2 roots of the symmetric quadratic (-x-1)^2 - x = 133 or x^2 + 2x + 1 - x = 133 or x^2 + x -132 = 0 So. x = (-1 +/- 23)/2 = (11, -12) = (a,b) or (b,a) by symmetry.
For a complete solution of the system you have to notice first the symmetry in a,b which shows that a= b is a possible case. Hence, there also two complex solutions together with solution pairs (a , b) ,(b,a) .
It is do easy equat 1 - equatio2 : ( a- b) * ( a+ b) = -( a-b,) therefore a=b cancell or a+b =,-1. B= - a-1 conclude a^2 +a +1= 133 conclude a =11 or -12. B= -12 or 11 thats all
I can't do anything, but I keep an eye on these types of problems and the solution comes quickly, in this case, as the values are different, it gets a little complicated, but the trick is that one value is positive and the other negative. a = - 12 b = + 11 144 - 11 = 133 121 - (-12) = 133 Bingo from Brazil
a^2 - b = b^2 -a a^2 + a = b^2 +b We are looking for two points on the parabola y = x^2 +x with same y-value. They must be symmetric to the lowest point with x coordinate -0.5 So a = -0.5 - c and b = -0.5 + c (-0-5 - c)^2 - (-0.5 + c) = 133 0.25 + c +c^2 +0.5 - c = 133 c^2 = 132.25 c = root (13225/100) = root (25*529/100) = 5 * 23 / 10 = 11.5 (or -11.5) So a = -12 and b = 11 (or opposite)
I am not sure if my solution is acceptable. My initial thoughts were a^2 + b = 133 -- (1) b^2 + a = 133 -- (2) (1) = a^2 + b = 121 + 12 Therefore a = 11 or -11 b = 12 However equation (2) shows that the vice versa. And hence one must be a negative therefore a = -11, b = 12 combinations would work for both equations.
Where are the other real roots? another solutions: (a,b) = (12.0434) [approximations for irrational numbers] (a,b) = (-11.0434) [approximations for irrational numbers] like a² - a = 133
Từ hệ phương trình đã cho, suy ra : a*2 - b=b*2 - a tương đương a*2 - b*2= - (a - b) (a+b)(a - b)= -(a - b) Chia hai vế cho (a - b),ta được: a+b= -1 hay -b=a+1 Thay vào a*2 - b=133 ta có a*2 +(a+1)=133 a*2 +a - 132=0 Giải phương trình trên thì đạt được hai cặp nghiệm: a=11 ;b= -12 a= -12 ;b=11
① +② get ③ the same ①+③ → a^2 + a = 132 a = 11 or -12 → ② if a =11 then b ^2 - 11 = 133 b = ±12 → ① b = 12 (X) b = -12 (O) if a = -12 then b^2 + 12 =133 b = ±11 → ① b = 11 (O) b = -11 (X) Answer： a =11 and b = -12 or a = -12 and b = 11
Once a+b+1 is determined to be zero, simply substitute b = -a−1 in first equation to get a²−(-a−1) = 133 a²+a−132 = 0 (a+12)(a−11) = 0 (a, b) ∈ {(-12, 11), (11, -12)}. Only one check is needed because the system is symmetric. Correction: first two equations were slightly wrong.
a² - b² + a - b = 0 (a - b)(a + b + 1) = 0 a = b => rejected b = - (a + 1) a² + a + 1 = 133 a² + a - 132 = 0 a = (-1 ± 23)/2 a = 11 => b = -12 a = -12 => b = 11
At the 154 seconds, you have violated PEMDAS, whereby, you added ((a-b) + (a-b) before multiplying (a+b) (a-b). This changes the answer. Please explain.
From the given equation we can infer that a² and b² are close to 133 and that perfect squares near this number are 121 and 144. Hence the value of a and b is either + or - 12 and + or - 11 or vice versa. By suitably substituting these nos. in the given equation we may get the answer in simple way.
Was it necessary to impose the condition a≠b in the question? It is clear that even if such a condition is not there a= b cannot be a solution since the LHS of each equation then becomes zero which is not equal to the RHS, 133.
