-1

Find the least degree polynomial equation with rational coefficients whose roots is given $\sqrt2-1$

My book has given the solution as

$x=(\sqrt2-1)$

$(x+1)^2 = 2$

I don't understand how this became the required polynomial equation.......I can't even determine if this has $(\sqrt2-1)$ even as a solution by looking at it. They have done the similar for obtaining $1+\sqrt2+\sqrt3$ as a root. For another polynomial least degree

Abhinav
  • 239
  • If you plug in $(x+1)^2 = 2$ the value of $x=\sqrt2-1$ you get $(\sqrt2-1+1)^2 = 2$ which is a tautology, thus it is a solution. But, plugging in $x = 1+\sqrt2+\sqrt3$ we get that it is definitly not a root. Perhaps you should edit your question. – yakobyd Jun 26 '18 at 10:53
  • They have just rearranged the equation. So you start with $x = \sqrt{2} - 1$, then add $1$ to both sides and square both sides to get $(x + 1)^2 = 2$. By expanding the left hand side and subtracting $2$ from both sides, you get a polynomial in $x$ with right hand side $0$ which means that $x$ is a root of that polynomial. – Bilbottom Jun 26 '18 at 10:56
  • @Bill walls Ah!I understood what you said.Basically x is the root of the polynomial. And we already have the value of x as √2-1 making √2-1 directly a root. – Abhinav Jun 26 '18 at 11:01

1 Answers1

3

Let’s do that step by step. Write

$$x=\sqrt{2}-1$$

This is equivalent to

$$x+1=\sqrt{2}$$

This means that $x$ is one of the two real numbers (the other one is $-1-\sqrt{2}$) such that

$$(x+1)^2=2$$

This rewrites as

$$x^2+2x-1=0$$

Is this the smallest degree with rational coefficients ? Yes because over the real numbers the smallest degree possible equation would be

$$x-\sqrt{2}+1=0$$

The coefficients cannot be rational because this would mean $\sqrt{2}$ is rational

marwalix
  • 16,773