# What are all solutions of equation f(-3x)=0 if the roots of equation f(x)=0 are -1 and -2?

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If -1 and 2 are roots, we can write:

x = -1 and x = 2.

Then x + 1 = 0 and x - 2 = 0

So the factors are x + 1 and x + 2.

And the simplest version of the original function would have been f(x) = (x + 1)(x + 2).

Replacing x with -3x gives us:

f(-3x) = (-3x + 1)(-3x + 2).

Setting these factors equal to 0 gives us:

-3x + 1 = 0 or -3x + 2 = 0

-3x = -1 or -3x = -2

x = 1/3 or x = 2/3

**The solution set is: {1/3, 2/3}.**

The roots of f(x) = 0 are -1 and -2. We can write f(x) = (x + 1)(x + 2)

f(-3x) = 0

=> (-3x + 1)(-3x + 2) = 0

The roots of this are x = 1/3 and x = 2/3

**The solutions of the equation are x = 1/3 and x = 2/3**

Since the equation f(x) = 0 has two roots, then the equation is a quadratic:

f(x) = (x+1)(x+2)

f(x) = x^2 + 3x + 2

f(-3x) = (-3x)^2 + 3*(-3x) + 2

f(-3x) = 9x^2 - 9x + 2

We'll apply quadratic formula:

x1 = [9+sqrt(81 - 72)]/18

x1 = (9+3)/18

x1 = 12/18

x1 = 2/3

x2 = (9-3)/18

x2 = 6/18

x2 = 1/3

Another way to solve the problem is to consider that any root of an equation, substituted within equation, verifies it.

f(-1) = 0

But f(-3x) = 0

f(-3x) = f(-3*(1/3)) = f(-1) = 0 => x = 1/3

f(-3x) = f(-3*(2/3)) = f(-2) = 0 => x = 2/3

**Therefore, the solutions of the quadratic equation f(-3x)=0 are {1/3 ; 2/3}.**