Find all the pairs (a,b) of real numbers such that whenever
α is a root of x^2 + ax + b = 0, α^2 - 2 is also a root of the equation.
Hello,
Rought method :
$\displaystyle x^2+ax+b$ can be factored into $\displaystyle (x-x_1)(x-x_2)=0$ where $\displaystyle x_1$ and $\displaystyle x_2$ are the roots of the equation.
--> $\displaystyle x^2+ax+b=(x-a)(x-(a^2-2))=(x-a)(x-a^2+2)$
Developping the RHS, we get : $\displaystyle x^2-ax-a^2x+2x+a^3-2a=x^2-x(a^2+a-2)+a^3-2a$
--> a^2+a-2=-a and a^3-2a=b
--------------------------
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
--------------------------
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) ----> $\displaystyle a_1=-1+ \sqrt{3}$ and $\displaystyle a_2=-1-\sqrt{3}$
And then calculate $\displaystyle b_1=a_1^3-2a_1$ and $\displaystyle b_2=a_2^3-2a_2$