# Spivak Calculus - Chapter 1 Question 4.6

In Spivak's Calculus, Chapter 1 Question 4.6:

Find all the numbers $x$ for which $x^2+x+1>2$

The chapter focuses on using the following properties of numbers to prove solutions are correct:

Based on those properties, I am able to perform the following algebra:

\begin{align} x^2 + x + 1 &> 2 & \text{Given}\\ x (x + 1) + 1 &> 2 & \text{P9}\\ x (x+1) &> 1 & \text{P3 P2 and Addition} \end{align}

And from there, I can note that:

\begin{align} x &\neq (x+1)^{-1}\\ x^{-1} &\neq (x+1)\\ \end{align}

By P6, because $x (x+1) > 1$ and $x (x+1) \neq 1$.

However, in his book Spivak is able to find the following:

\begin{align} x &> \frac{-1+\sqrt{5}}{2} \text{ or}\\ x &< \frac{-1-\sqrt{5}}{2} \end{align}

How does he come to that conclusion using only the properties listed above?

• $x^2+x+1>2\iff x^2+x-1>0\iff (x-\frac{-1+\sqrt{5}}{2})(x-\frac{-1-\sqrt{5}}{2}>0$ and a product of two non-zero reals is positive if and only if they have the same sign. – Jorge Fernández Hidalgo Aug 1 '16 at 21:05
• To be more specific, either prove the quadratic formula (which is in one of the exercises) or explicitly complete the square in this example. – Ted Shifrin Aug 1 '16 at 21:06

\begin{align} x^2+x+1&>2 & \text{Given}\\ x^2+x+1+0&>2+0 & \text{By Addition}\\ x^2+x+1+0&>2 & \text{By P2}\\ x^2+x+0+1&>2 & \text{By P4}\\ x^2+x+\left( \frac{1}{2} \right)^2+(-1)\left( \frac{1}{2} \right)^2+1 &>2 & \text{By P3}\\ \left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right)+(-1)\left( \frac{1}{2} \right)^2+1 &>2 & \text{By P9}\\ \left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right)+ (-1)\left( \frac{1}{4} \right) + 1 &> 2 & \text{By Multiplication}\\ \left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right) &> \left( \frac{5}{4} \right) & \text{By Addition, P3, and P2}\\ \end{align}