a² = b + 183 b² = a + 183 ------------------------------------------------subtraction a² - b² = (b + 183) - (a + 183) a² - b² = b + 183 - a - 183 a² - b² = - a + b a² - b² + a - b = 0 (a + b).(a - b) + (a - b) = 0 (a - b).[(a + b) + 1] = 0 (a - b).(a + b + 1) = 0 → given: a ≠ b → a - b ≠ 0 a + b + 1 = 0 b = - 1 - a Restart: a² = b + 183 → where: b = - 1 - a a² = - 1 - a + 183 a² + a - 182 = 0 Δ = (1)² - (4 * - 182) = 729 = 27² a = (- 1 ± 27)/2 First case: a = (- 1 + 27)/2 a = 13 → recall: b = - 1 - a b = - 14 Second case: a = (- 1 - 27)/2 a = - 14 → recall: b = - 1 - a b = 13
a² = b + 183
b² = a + 183
------------------------------------------------subtraction
a² - b² = (b + 183) - (a + 183)
a² - b² = b + 183 - a - 183
a² - b² = - a + b
a² - b² + a - b = 0
(a + b).(a - b) + (a - b) = 0
(a - b).[(a + b) + 1] = 0
(a - b).(a + b + 1) = 0 → given: a ≠ b → a - b ≠ 0
a + b + 1 = 0
b = - 1 - a
Restart:
a² = b + 183 → where: b = - 1 - a
a² = - 1 - a + 183
a² + a - 182 = 0
Δ = (1)² - (4 * - 182) = 729 = 27²
a = (- 1 ± 27)/2
First case: a = (- 1 + 27)/2
a = 13 → recall: b = - 1 - a
b = - 14
Second case: a = (- 1 - 27)/2
a = - 14 → recall: b = - 1 - a
b = 13
(a+b)(a-b)+(a-b)=0 then how do you get (a_b)(a+b)+1=0? why not (a+b)+1=0?
It's (a-b)[(a+b)+1]=0 ,
not (a-b)(a+b)+1=0.
Hence, (a-b)[(a+b)+1]=0
and (a+b)+1=0. Are the same
Since, (a-b)≠0. {a≠b}.
@@thomasslemmer6527
感謝。
1) a^2 = b + 183
2) b^2 = a + 183
1) - 2) a^2 - b^2 = b - a
a + b = -1
(b + 1)^2 = b + 183
b^2 + b - 182 = 0
b = (-1 +/- √729)/2 = (-1 +/- 27)/2 = 13, -14
(a, b) = (-14, 13), (13, -14)
Correct. This is the straight forward method
@ 🙂
x=（1+√733）/2,(1−√733)/2
Does it mean more money if you make your videos longer unnecessarily? Some of tye,steps you take are just silly.
I make longer videos for deeper understanding. Crash course!
Спасибо за терпеливое отношение к своим ученикам❤@@superacademy247
Silly for teachers but not for learning pupils
When ais 12 b is -39 . When bis 12 a is -39.