User:Ihri/Sandbox

Theorem
Let $a$ and $b$ be (strictly) positive real numbers such that $a^2 - b > 0$.

Then:

Proof
We are given that $a>0$ and $b>0$.

Then:
 * $a + \sqrt b > 0$

and so $\displaystyle \sqrt {a + \sqrt b}$ is defined on the real numbers.

Consider the quadratic equation:


 * $z^2 - a z + \dfrac b 4 = 0$

with discriminant $(- a) ^ 2 - 4 \dfrac b 4 = a^2 - b$ which is a (strictly) positive real number (a given).

Let $x, y$ be the solutions of the above equation.

From Solution to Quadratic Equation, :

with $x > y$. Note $x > 0$ because $ a > 0$.

By Sum of Roots of Quadratic Equation, $x$ and $y$ satisfy:


 * $x + y = a$

By Product of Roots of Quadratic Equation and the conditions $x > 0, b >0$ we get:


 * $x y = \dfrac b 4 > 0 \implies y > 0$

which implies $\displaystyle \sqrt {x}$ and $\displaystyle \sqrt {y}$ are defined on the real numbers.

Furthermore, from $x y = \frac b 4$ we get:

Finally,

{{eqn | l = \sqrt {a + \sqrt b}     | r = \sqrt {x + y + 2 * \sqrt xy}} | c = }}

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Let $\displaystyle \sqrt {a + \sqrt b} = \sqrt x + \sqrt y$ where $x, y$ are (strictly) positive real numbers.

Squaring both sides gives:

Set $x + y = a$ and $\sqrt b = 2 \sqrt {x y}$

From $\sqrt b = 2 \sqrt {x y}$ we get:

By Viète's Formulas, $x$ and $y$ are solutions to the quadratic equation:


 * $z^2 - a z + \dfrac b 4 = 0$

From Solution to Quadratic Equation:


 * $z_{1, 2} = \dfrac {a \pm \sqrt {a^2 - b} } 2$

where $a^2 - b > 0$ (which is a given)



Subsituting into $\displaystyle \sqrt {a + \sqrt b} = \sqrt x + \sqrt y$: