Are the a's all meant to be the same variable?
The first step either way would be to recognize that α(α^2 - 2) = b and (α) + (α^2 - 2) = -a.
Hello,
Rought method :
can be factored into where and are the roots of the equation.
-->
Developping the RHS, we get :
--> a^2+a-2=-a and a^3-2a=b
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Other method that will end up at the same point as above
In an equation x²+mx+n=0, the sum of the roots is -m and the product of the roots is n.
Here, we have x²+ax+b=0 and roots a and a²-2.
Therefore (a²+a-2)=-a and a^3-2a=b
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a²+a-2=-a ------> a²+2a-2=0.
Completing the square :
a²+2a+1-3=0 ------> (a+1)²=3 -----> a+1=+ or - sqrt(3) ----> and
And then calculate and