Math Olympiad | A Nice Algebra Problem | VIJAY Maths

Also Watch our Most Viral Interesting Math Olympiad Problem:
• Math Olympiad | A Nice...
Subscribe to our channel and press the bell icon 🔔 for daily Brainstorming Math videos →
/ @vijaymaths5483
*****************************************************************************
#exponentialproblems #matholympiad #maths #algebra

[@key_board_x]

ab/bc = 1/2 a/c = 1/2 2a = c → given: ac = 3 → c = 3/a 2a = 3/a 2a² = 3 a² = 3/2 bc/ac = 2/3 b/a = 2/3 b = (2/3) * a b² = (4/9) * a² b² = (4/9) * (3/2) b² = 12/18 b² = 2/3 Recall: 2a = c c = 2a c² = 4a² c² = 4 * (3/2) c² = 6 (a + b + c)² = (a + b + c).(a + b + c) (a + b + c)² = a² + ab + ac + ab + b² + bc + ac + bc + c² (a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc (a + b + c)² = a² + b² + c² + 2.(ab + ac + bc) (a + b + c)² = (3/2) + (2/3) + 6 + 2.(1 + 2 + 3) (a + b + c)² = (13/6) + 18 (a + b + c)² = 121/6 a + b + c = ± √(121/6) a + b + c = ± √(121)/(√6) a + b + c = ± 11/(√6) a + b + c = ± (11√6)/6

[@artworldpro4971]

(ab)(bc)(ca)=1×2×3 a²b²c²=6 abc=√6 a=(√6)/bc a=√6/2 b=(√6)/ac b=√6/3 c=(√6)/ab c=√6/1=√6 a+b+c=√6/2+√6/3+√6 =(3√6+2√6+6√6)/6=11√6/6

[@RealQinnMalloryu4]

(ab ➖1ab+1) (bc ➖1bc+1) (ca ➖2 ca+1).

[@bly9925]

bc=2=2ab gives c=2a. 2a^2=3 gives a=sqrt(3/2), which in turn gives b=sqrt(2/3). Finally, c=sqrt(6). Hence, a+b+c=sqrt(3/2)+sqrt(2/3)+sqrt(6)=(2+3+6)/sqrt(6)=11/sqrt(6).

[@jaye-r3r]

the previous handwriting with a marker is much better than the new one.

[@vijaymaths5483]

Thank you 😀

[@pas6295]

It is similar to earlier one. Now you have. ab=1. Then bc=2. And then came=3. Multiplying the first one and the second one you get a^2(bc)=a^2(2)and this is =1×[e-mail] 2a^2=2 So ^2=1. Like that you can get b value and value for c. Also. By adding you get the answer.

[@raghvendrasingh1289]

Multiplying all equations and taking square root abc = √6 Now a+b+c = abc (1/bc+1/ca+1/ab) = √6 (1/2+1/3+1) = √6×11/6 = 11√6/6

[@souzasilva5471]

Porque você não considerou a soma negativa? Why didn't you consider the sum negative?

[@Quest3669]

Shorts.. a/c= 1/2; then a^2= 3/2 or a= √3/√2 then b= √2/√3; c= √6 a+b+c=11/ √ 6

[@davez8816]

Good but you have forgotten the negative value a+b+c=-11sqrt(6)/6

[@sandytanner9333]

a=√(3/2) b=√(2/3) c=√6 a+b+c= (1/√6)(3+2+6) a+b+c= 11/√6

from Morocco thank you.....at mn 3-04 we had a=1/b we can replace and deduce that abc=a.(1/a)c=sq6....then c=sq6.....then get all values of b=2/sq6 and a=3/sq6 then compute theire summ

[@Rocio62154]

SENSATIONAL!