Olympiad Mathematics leading to quadratic equation.
This is beautiful solved #maths #viral #mathematics #mathstricks #exponential
Жүктеу.....
Пікірлер: 3
@prollysine19 күн бұрын
let u=sqrt(x) , x=u^2 , u^2+u-8=0 , u= (-1+sqrt(1+32))/2 -> x^(1/2)= (-1+sqrt(1+32))/2 , x=(-1+sqrt(1+32)^2/4 , x^(1/2)= ~ 2.37228 , x=~ 5.62722 , 8-5.62722=2.37228 , same , OK , x^(1/2)= (-1-sqrt(1+32))/2 , x^(1/2)=~ -3.37228 , x=(-1-sqrt(1+32))^2/4 , x=11.3723 , 8-11.3723= -3.37228 , same , OK ,
@Baltie320 күн бұрын
It's not the olympiad level.
@rvqx19 күн бұрын
I think you are wrong. I found only 1 solution: x= 8.5 - 0.5V33 . Then i checked graphically and still 1 solution. Then i calculated your first x : x=11.37228 In the original equasion : V11.37228 = 8 - 11.37228 is not correct. My solution: let y=Vx then y=8 - y**2 This gives only 1 positive solution for y and thus 1 solution for x.
Пікірлер: 3
let u=sqrt(x) , x=u^2 , u^2+u-8=0 , u= (-1+sqrt(1+32))/2 -> x^(1/2)= (-1+sqrt(1+32))/2 , x=(-1+sqrt(1+32)^2/4 , x^(1/2)= ~ 2.37228 , x=~ 5.62722 , 8-5.62722=2.37228 , same , OK , x^(1/2)= (-1-sqrt(1+32))/2 , x^(1/2)=~ -3.37228 , x=(-1-sqrt(1+32))^2/4 , x=11.3723 , 8-11.3723= -3.37228 , same , OK ,
It's not the olympiad level.
I think you are wrong. I found only 1 solution: x= 8.5 - 0.5V33 . Then i checked graphically and still 1 solution. Then i calculated your first x : x=11.37228 In the original equasion : V11.37228 = 8 - 11.37228 is not correct. My solution: let y=Vx then y=8 - y**2 This gives only 1 positive solution for y and thus 1 solution for x.