Show that there exists a root of the equation
$ x^2-x-1= \frac{1}{x+1} $
I don't know where to start. I need hints.
$\endgroup$ 33 Answers
$\begingroup$Hint: Define $$f(x)=x^2-x-1-\frac{1}{x+1}$$
that function is continuous in $\Bbb{R}-\{-1\}$. Note that $f(1)<0$ and $f(2)>0$. What the intermediate value theorem says?
$\endgroup$ $\begingroup$I presume you are looking to show that the the polynomial has a real root. The equation clearly has at least one complex root by the fundamental theorem of algebra.
Any roots of your equation will be equivalent to the roots of the polynomial $$P(x)=(x+1)(x^2-x-1)-1$$ which is a real cubic equation. Real cubic equations must have at least one real root, by the conjugate root theorem (or by IVT)
$\endgroup$ $\begingroup$Multiply both sides by $(x+1)$, you have:
$$(x^2-x-1)(x+1)=x^3-x^2-x+x^2-x-1=x^3-2x-1=1$$
$$x^3-2x-2=0$$
You can check that $x^3-2x-2 \to -\infty$ when $x \to -\infty$ and $x^3-2x-2 \to \infty$ when $x \to \infty$, so there exists root of this equation (you can check that it is not $x=-1$).
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