Root of the equation?

**fardeen_gen** · July 19th 2008, 07:32 AM

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.

**Henderson** · July 19th 2008, 08:22 AM

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.

**Moo** · July 19th 2008, 08:24 AM

Hello,

Originally Posted by fardeen_gen

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.

Rought method :
$x^2+ax+b$ can be factored into $(x-x_1)(x-x_2)=0$ where $x_1$ and $x_2$ are the roots of the equation.

--> $x^2+ax+b=(x-a)(x-(a^2-2))=(x-a)(x-a^2+2)$

Developping the RHS, we get : $x^2-ax-a^2x+2x+a^3-2a=x^2-x(a^2+a-2)+a^3-2a$

--> 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) ----> $a_1=-1+ \sqrt{3}$ and $a_2=-1-\sqrt{3}$

And then calculate $b_1=a_1^3-2a_1$ and $b_2=a_2^3-2a_2$

**fardeen_gen** · July 20th 2008, 05:40 AM

So what are the pairs of numbers?

**Moo** · July 20th 2008, 08:12 AM

It's all computation...

$b_1=a_1^3-2a_1=(-1+ \sqrt{3})^3-2(-1+ \sqrt{3})=\dots=4 \sqrt{3}-8$

So the first couple is $(-1+ \sqrt{3}~,~4 \sqrt{3}-8)$

Do the same with the second

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