# Convergent Real Sequence/Examples/x (n+1) = x n^2 + k/Lemma 2

## Example of Convergent Real Sequence

Let $a$ and $b$ be the roots of the quadratic equation:

$(1): \quad x^2 - x + k = 0$

Let:

$0 < k < \dfrac 1 4$

Then $a$ and $b$ are both strictly positive real numbers.

## Proof

First we investigate the consequences of the condition $k < \dfrac 1 4$.

In order for the quadratic equation $a x^2 + b x + c$ to have real roots, its discriminant $b^2 - 4 a c$ needs to be strictly positive.

The discriminant $D$ of $(1)$ is:

 $\ds D$ $=$ $\ds \paren {-1}^2 - 4 \times 1 \times k$ $\ds$ $=$ $\ds 1 - 4 k$

Thus:

 $\ds k$ $<$ $\ds \dfrac 1 4$ $\ds \leadsto \ \$ $\ds 1 - 4 k$ $>$ $\ds 0$

and so when $k < \dfrac 1 4$, $(1)$ has real roots.

$\Box$

Next we investigate the consequences of the condition $0 < k$.

 $\ds x$ $=$ $\ds \dfrac {-b \pm \sqrt {b^2 - 4 a c} } {2 a}$ where $b = -1$, $a = 1$, $c = k$ in $(1)$ $\ds$ $=$ $\ds \dfrac {-\paren {-1} \pm \sqrt {\paren {-1}^2 - 4 \times 1 \times k} } {2 \times 1}$ $\ds$ $=$ $\ds \dfrac {1 \pm \sqrt {1 - 4 k} } 2$

We have that:

 $\ds 0$ $<$ $\ds k$ $\ds \leadsto \ \$ $\ds 1$ $>$ $\ds 1 - 4 k$ $\ds \leadsto \ \$ $\ds 1$ $>$ $\ds +\sqrt {1 - 4 k}$ $\ds -1$ $<$ $\ds -\sqrt {1 - 4 k}$ $\ds \leadsto \ \$ $\ds \dfrac {1 \pm \sqrt {1 - 4 k} } 2$ $>$ $\ds 0$

That is, when $0 < k$ both roots of $(1)$ are strictly positive.

$\Box$

Hence when $0 < k < \dfrac 1 4$, both roots of $(1)$ are strictly positive real numbers.

$\blacksquare$