A Very Nice Math Olympiad Problem | Solve for all values of x? | Algebra Equation

In this video, I'll be showing you step by step on how to solve this Olympiad Maths Algebra problem using a simple trick.

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[larry]

I understand about complex numbers, but it’s still a mystery to me that you can have a complex solution on the left-hand side, inside the parentheses, that does not equal 2, be equal to the 2 that is raised to 6th power on the right hand side.

[yifengxiao]

exponents equal, then bases should be equal. since the power is 6, so x+3=±2. thus, x=-1, and x=-5.

[Limited_Light]

8:05 0 = x^2 + 8x + 19
= x^2 + 2 * 4 * x + 19
= x^2 + 2 * 4 * x + 16 + 3
= x^2 + 2 * 4 * x + 4^2 + 3
= (x + 4)^2 + 3
===> (x + 4)^2 = -3
===> x + 4 = [plus or minus] sqrt(-3)
===> x = -4 [plus or minus] i * sqrt(3).

[kareolaussen]

x=-3+2*exp(iπk/3), k=-2, -1, 0, 1, 2, 3.

[andretewem]

Why do not use the polar form ?
1. z = x + 3
2. set z = (|z|, arg(z))
==> z^6 = (|z|^6, 6.arg(z))
3. 2^6 = (2^6, 0 + 2k.pi)
Then (x + 3)^6 = 2^6 <==>
|z|^6 = 2^6 and 6.arg(z) = 2k.pi
<==> |z| = 2 and arg(z) = k.pi /3
with k € [0 ; 1 ; ... ; 5].
The solutions in "x" are - 2, 2,
1 +/- i.sqrt(3),
-1 +/- i.sqrt(3).

The solutions in "z" are -5, -1,
-2 +/ i.sqrt(3),
-4 +/- i.sqrt(3)

[artursierant]

divide both sides by 2^6, you get 6 roots of 1

[PawełDąbrowski-nw]

(x+3)⁶-2⁶ = 0
((x+3)³)²-(2³)² = 0
((x+3)³+2³)((x+3)³-2³) = 0
(x+3+2)(x²+6x+9-2x+10) = 0

Lets analyze this first and then we'll proceed next part:

X=-5

x²+4x+19 = 0
x²+4x+4 = -15
(x+2)² = -15
x=-2±i√15

There are next 2 answers to our problem. Let's check what we've left behind.

(x+3)³-2³ = 0

(x+1)(x²+6x+9+2x+6+4) = 0
x= -1

x²+8x+19 = 0
x²+8x+16 = -3
(x+4)² = -3
x= -4±i√3

[TonyFisher-lohh]

Incompleteiy specified problem. In the realm of reals, it has 2 solutions (-1 and -5). In the realm of complex there are more.

[heribertoayalareyes]

@nelly. Ecuac grado 6 tiene 6 raíces!!