6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
4
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.