Algebraic Solution to $z^2 = az + b^2$

Question

In a book about the history of invisible numbers, the author writes:

$\frac 1 2 a + \sqrt {(\frac a 2)^2 + b^2}$ is the solution to $z^2 = az + b^2 $

Where is this coming from? I could not find a way to work it out through the quadratic formula.

Answer 1

Let z = (1/2)a + √((a/2)^2 + b^2) and
z-a = -(1/2)a + √((a/2)^2 + b^2)
Then using the formula (-x+y)(x+y) = -x^2+y^2,
we get z(z-a) = -(a/2)^2 + ((a/2)^2 + b^2) = b^2.

Answer 2

The solution to $z^2-az-b^2$ is $$z = \frac{a\pm\sqrt{a^2+4b^2}}{2} = \frac{1}{2}a \pm \frac{1}{2}\sqrt{a^2+4b^2} = \frac{1}{2}a \pm \sqrt{\frac{a^2+4b^2}{4}} = \frac{1}{2}a \pm \sqrt{\left(\frac{a}{2}\right)^2 + b^2}.$$

Answer 3

It is coming from the Quadratic formula! If you have an equation in the form $ax^2+bx+c=0,\:a\neq0$ (Quadratic),then you can use this equation to solve for x: $$x=\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$$

Turn the equation,$z^2=az+b^2$, into a Quadratic and then apply the formula.

$$z^2=az+b^2$$ $$-z^2+az+b^2=0$$ If you don't see it, substitute $x=z$,$c=b^2$,$a=b$! You get $-x^2+bx+c=0$ where $a=-1$! But's let's just substitute directly into the quadratic formula.

We know: $a=-1$,$b=a$,$c=b^2$, $z=x$. Now plug it into the formula! $$x=\frac{-a\pm\sqrt{\left(a\right)^2-4\cdot -1\cdot b^2}}{2\cdot -1}$$ Split this into two fractions. $$x=\frac{-a}{-2\:}\pm \:\:\:\frac{\:\sqrt{\left(a\right)^2\cdot \:\:\:+4\:\:b^2}}{2}=\frac{-a}{-2\:}\pm \:\:\:\:\frac{\:\sqrt{\left(a\right)^2\cdot \:\:\:\:+4\:\:b^2}}{\sqrt{4}}$$

$$x=\frac{1}{2}a\:\pm \:\:\:\:\:\sqrt{\left(\frac{a}{2}\right)^2\cdot \:\:\:\:+4\:\:b^2}$$