If an area is contained by a rational (straight line) and a fourth binomial (straight line) then the square root of the area is the irrational (straight line which is) called major. * For let the area $AC$ be contained by the rational (straight line) $AB$ and the fourth binomial (straight line) $AD$, which has been divided into its (component) terms at $E$, of which let $AE$ be the greater. * I say that the square root of $AC$ is the irrational (straight line which is) called major.
If the rational straight line has unit length then this proposition states that the square root of a fourth binomial straight line is a major straight line: i.e., a fourth binomial straight line has a length \[\alpha\,\left(1+\frac 1{\sqrt{1+\beta}}\right)\] whose square root can be written \[\rho\,\sqrt{\frac 12+\frac\delta{2\sqrt{1+\delta^{\,2}}}}+\rho\,\sqrt{\frac 12 -\frac\delta{2\sqrt{1+\delta^{\,2}}}},\] where \[\rho=\sqrt{\alpha}\quad\text{ and }\quad\delta^{\,2}=\beta.\]
This is the length of a major straight line (see [Prop. 10.39]), since $\rho, \delta,\alpha,\beta$ are all positive rational numbers.
Proofs: 1