Solve in $\mathbb{Q}$ the equation $x^2-(\sqrt{2}+1)x+\sqrt{2}=0$.

Question

Solve in $\mathbb{Q}$ the equation $x^2-(\sqrt{2}+1)x+\sqrt{2}=0$

Somebody can help me? I dont remember how to do.

Answer 1

Using Quadratic Formula

If $ax^2+bx+c=0\;,$ Then $\displaystyle x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here $a = 1$ and $b=-\left(\sqrt{2}+1\right)$ and $c=\sqrt{2}$

So $$\displaystyle x = \frac{(\sqrt{2}+1)\pm \sqrt{2+1+2\sqrt{2}-4\sqrt{2}}}{2}$$

So $$\displaystyle x = \frac{(\sqrt{2}+1)\pm (\sqrt{2}-1)}{2} = \sqrt{2}\;\;,1$$

Answer 2

$$x^2 - \left(\sqrt 2 + 1\right) x + \sqrt 2 = \left(x-\sqrt 2\right)\left(x - 1\right) = 0$$

Since your solution $x^* \in \mathbb{Q},\ \ x^* = \dots$