The whole point of most of these so-called Olympiad problems is that, yes, you could take the square root of both sides and get the real roots, but many of them have more than just real roots. They have complex roots. The fundamental theorem of algebra says that if you have an equation, where the leading coefficient is a certain number then you must have at least that many number of solutions, although some of them may be complex and not real solutions.
(x+3)²=3² =>|x+3|=|3|
|x+3|=x+3 if x>3
|x+3|=-(x+3) if x x+3=3 or x+3=-3
x=3-3 or x=-3-3
x=0 or x=-6
Thanks
@@PhilCoolMath the square root of a number x² is equal to the absolute value of x (|x|) generally.
|x|=x if x>0
|x|=-x if x
Why not take the square root of both sides. In a single minute you will get 0 and -6
Jah that works, it's even the simplest or obvious way most people will solve it
i agree...hardly an Olympiad problem.
The whole point of most of these so-called Olympiad problems is that, yes, you could take the square root of both sides and get the real roots, but many of them have more than just real roots. They have complex roots. The fundamental theorem of algebra says that if you have an equation, where the leading coefficient is a certain number then you must have at least that many number of solutions, although some of them may be complex and not real solutions.
(x + 3)² = 3²
Ix+3I = 3
x+3 = 3 or x+3 = - 3
x=0 or x = - 6
Pourquoi faire simp,e quand on peut faire compliqué. (x+3)²=3² égale x+3=3 ou x+3=-3
(x+3)^2=3^2 ={{{{{{{{{ x+5= 5 X=5-5=1 X=1
5-5=0
That's one solution, what about the other?
X is 0
-6, 3.
(x+3)²=3² =>|x+3|=|3|
|x+3|=x+3 if x>3
|x+3|=-(x+3) if x x+3=3 or x+3=-3
x=3-3 or x=-3-3
x=0 or x=-6
Thanks