I have got a question which looks like follow:

Show that the equation $\sqrt{ax+\alpha}+\sqrt{bx+\beta}+\sqrt{cx+\gamma}=0$ reduces to a simple equation if $\sqrt{a}\pm\sqrt{b}\pm\sqrt{c}=0$.

I am totally confused and don't even know from where should I start.

Side note: I don't know what is a simple equation (I think it is something which is not filthy like given equation).

Any hint/suggestion is heartily welcome. Thanks.

  • $\begingroup$ What are the value ranges ? $\endgroup$ – user90369 Dec 9 '16 at 18:04
  • $\begingroup$ I won't lie, the book has no mention of range. $\endgroup$ – Vidyanshu Mishra Dec 9 '16 at 18:05
  • 2
    $\begingroup$ Possibly of help is my answer to Rationalizing radicals, in particular where I show that $(a+b+c)(a+b-c)(a-b+c)(a-b-c)$ is equal to $\left(a^2 - b^2\right)^2 \; - \; 2c^2\left(a^2 + b^2\right) \; + \; c^4.$ $\endgroup$ – Dave L. Renfro Dec 9 '16 at 18:06
  • $\begingroup$ What about $x$? Is real? $\endgroup$ – Ricardo Largaespada Dec 10 '16 at 5:35
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    $\begingroup$ @Semiclassical Since no range for $a,b,c\alpha,\beta,\gamma$ is give I thing there is a massive possibility that these can be any numbers (even complex). $\endgroup$ – Vidyanshu Mishra Dec 14 '16 at 17:00

This question make more sense when we don't limit outselves to real numbers.

For any $n = d + 1 > 1$ and $u = (u_0, u_1, \ldots, u_d) \in \mathbb{C}^n$, consider following product

$$\Lambda(u_0,\ldots,u_d) = \prod_{(\epsilon_1,\ldots,\epsilon_d) \in \{ \pm 1 \}^d} \left( \sqrt{u_0} + \sum_{k=1}^d \epsilon_k \sqrt{u_k} \right)\tag{*1} $$ We can expand $\Lambda(\cdots)$ to a homogeneous polynomial in $\sqrt{u_k}$ of degree $2^d$

$$\Lambda(u_1,\ldots,u_d) = \sum_{(e_0,\ldots,e_d)\in \mathbb{N}^n} A_{e_0,\ldots,e_d} \prod_{k=0}^d\sqrt{u_k}^{e_k} \tag{*2} $$ whose coefficients $A_{e_0,\ldots,e_d} \in \mathbb{Z}$ and vanish unless $e_0 + \ldots + e_d = 2^d$.

Consider the effect of flipping the sign of $\epsilon_\ell$ for some $\ell \ge 1$ in $\Lambda(\cdots)$.

  • In $(*1)$, this rearrange the order of product but leaves the value of $\Lambda(\ldots)$ untouched.
  • In $(*2)$, the coefficient $A_{e_0,\ldots,e_d}$ picks up a factor $(-1)^{e_\ell}$.

Since the value of product doesn't change, $A_{e_0,\ldots,e_d}$ vanishes unless $e_\ell$ is even. Since this is true for every $\ell \ge 1$ and $A_{e_0,\ldots,e_d}$ vanishes unless $e_0 + \cdots + e_d = 2^d$, $A_{e_0,\ldots,e_d}$ also vanishes unless $e_0$ is even. This implies in expansion $(*2)$, all square roots get completed.

As a result, $\Lambda(\cdots)$ is a homogeneous polynomial in $u_0,\ldots, u_d$ of degree $2^{d-1}$:

$$\Lambda(u_1,\ldots,u_d) = \sum_{(e_0,\ldots,e_d)\in \mathbb{N}^n} B_{e_0,\ldots,e_d} \prod_{k=0}^du_k^{e_k} \tag{*3} $$ whose coefficients $B_{e_0,\ldots,e_d} \in \mathbb{Z}$ and vanish unless $e_0 + \ldots + e_d = 2^{d-1}$.

If $\sqrt{u_0} \pm \sqrt{u_1} \pm \cdots \pm \sqrt{u_d} = 0$ for any choice of sign of the square roots, then by construction, $u_0, \ldots, u_d$ need to satisfy the polynomial equation $\Lambda(u_0,\ldots,u_d) = 0$.

For the problem at hand, take $n = 3$ and substitute $(u_0,u_1,u_2)$ by $ (ax + \alpha, bx+\beta, cx+\gamma)$.
The equation $\sqrt{a x + \alpha} \pm \sqrt{b x + \beta} \pm \sqrt{cx + \gamma} = 0$ leads to a homogeneous polynomial equation in $ax, bx, cx, \alpha, \beta, \gamma$ of degree $2$:

$$\Lambda(ax+\alpha, bx+\beta, cx+\gamma) = 0$$

Expand this polynomial out against $x$, we obtain a quadratic equation in $x$:

$$C(\cdots) x^2 + D(\cdots) x + E(\cdots) = 0$$

It is easy to see the coefficients $C(\cdots)$ only depends on $(a,b,c)$. By setting $x$ to $1$ and $\alpha, \beta, \gamma$ to $0$, we find $C(\cdots) = \Lambda(a,b,c)$. By setting $x$ to $0$, we find $E(\cdots) = \Lambda(\alpha,\beta,\gamma)$. This leads to a equation of the form:

$$\Lambda(a,b,c)x^2 + D(\cdots)x + \Lambda(\alpha,\beta,\gamma) = 0$$

Now it comes to the mysterious condition $\sqrt{a} \pm \sqrt{b} \pm \sqrt{c} = 0$. When this condition is fulfilled, $\Lambda(a,b,c) = 0$. Above equation simplifies to a linear equation in $x$.

$$D(\cdots)x + \Lambda(\alpha,\beta,\gamma) = 0$$

We can determine the last unknown coefficient $D(\cdots)$ by setting $x$ to $1$. At the end, we have

When $\sqrt{a} \pm \sqrt{b} \pm \sqrt{c} = 0$, then the equation $\sqrt{ax+\alpha} \pm \sqrt{bx+\beta} \pm \sqrt{cx+\gamma} = 0$ leads to a linear equation in $x$: $$\Lambda(a+\alpha,b+\beta,c+\gamma)x + \Lambda(\alpha,\beta,\gamma)(1-x) = 0$$ where $$\Lambda(u,v,w) = u^2 + v^2 + w^2 - 2(uv+vw+uw)$$

In certain sense, one can argue this equation is simple because it's dependence on $x$ is linear. Unlike the general case where $\Lambda(a,b,c) \ne 0$, the solution for $x$ no longer involves any radicals. Whether one agree this is simple is up to one's own judgement. To be honest, I don't.

  • $\begingroup$ Very thorough answer (though a bit hard to follow at times). +1 $\endgroup$ – Fimpellizieri Dec 15 '16 at 5:32

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