# Given a,b the roots of equation x^2-3mx+18=0, what is m if a^3+b^3>=18m^2? (don't solve equation x^2-3mx+18=0)

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### 1 Answer

The problem specifies to you not to solve the given quadratic equation to evaluate the roots `a` and `b` , hence, you should replace `a` and `b` for `x` in the given equation, such that:

`a^2 - 3ma + 18 = 0 => a^3 - 3ma^2 + 18a = 0`

`b^2 - 3mb + 18 = 0 => b^3 - 3mb^2 + 18b = 0`

You need to perform the addition of the equations, such that:

`a^3 + b^3 - 3m(a^2 + b^2) + 18(a + b) = 0`

Keeping `a^3 + b^3` to the left side, yields:

`a^3 + b^3 = 3m(a^2 + b^2) - 18(a + b)`

Using Vieta's relations, yields:

`a + b = -(-3m)/1 => a + b = 3m`

`a*b = 18/1 = 18`

`a^2 + b^2 = (a + b)^2 - 2ab => a^2 + b^2 = (3m)^2 - 2*18`

`a^2 + b^2 = 9m^2 - 36`

Replacing `9m^2 - 36` for `a^2 + b^2` and `3m` for `a + b` , yields:

`a^3 + b^3 = 3m(9m^2 - 36) - 18*3m`

You need to solve for m the inequality `a^3+b^3>=18m^2` , hence, you need to replace `3m(9m^2 - 36) - 18*3m` for `a^3+b^3` , such that:

`3m(9m^2 - 36) - 18*3m >= 18m^2`

Factoring out `3m` to the left side, yields:

`3m(9m^2 - 36 - 18) >= 18m^2`

Moving all terms to one side, yields:

`3m(9m^2 - 54) - 18m^2 >= 0`

You need to factor out 3m, such that:

`3m(9m^2 - 6m - 54) >= 0`

Using the positive product rule, yields that the factors need to be both positive or both negative, such that:

`{(3m>=0),(9m^2 - 6m - 54 >= 0):}`

or

`{(3m<=0),(9m^2 - 6m - 54 <= 0):}`

Considering the first situation, yields:

`{(3m>=0),(9m^2 - 6m - 54 >= 0):} => {(3m=0),(3m^2 - 2m - 18 =0):} => {(m = 0),(m_(1,2) = (2+-sqrt(4 + 216))/6):} => m in (-oo,(1-sqrt55)/3)U((1+sqrt55)/3,oo) nn (0,oo) => m in ((1+sqrt55)/3,oo)`

Considering the next situation, yields:

`m in (-oo,0) nn ((1-sqrt55)/3,(1+sqrt55)/3) => m in ((1-sqrt55)/3,0)`

**Hence, evaluating m in both situations, yields either `m in ((1+sqrt55)/3,oo)` , or **`m in ((1-sqrt55)/3,0).`

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