Friday, April 10, 2015

Evaluate $abcd$

Let $a, b, c, d$ be real numbers such that [MATH]a=\sqrt{4-\sqrt{5-a}}[/MATH], [MATH]b=\sqrt{4+\sqrt{5-b}}[/MATH], [MATH]c=\sqrt{4-\sqrt{5+c}}[/MATH] and [MATH]d=\sqrt{4+\sqrt{5+d}}[/MATH]. Calculate $abcd$.

This is one very intriguing problem because I am sure your instinct will tell you, No, please don't multiply all the given four expression out! That is a surest way to get into a big mess if you do that, and there got to be an easy way out for this problem.

Indeed, you're absolutely correct because if we multiply all of the four equations out, we obtain:

$abcd=\sqrt{(4-\sqrt{5-a})(4+\sqrt{5-b})(4-\sqrt{5+c})(4+\sqrt{5+d})}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,=\sqrt{(16+4(\sqrt{5-b}-\sqrt{5-a})-\sqrt{(5-a)(5-b)})(16+4(\sqrt{5-d}-\sqrt{5-c})-\sqrt{(5-c)(5-d)})}$

You would know by now that there is no use to continue expanding the expressions inside the square root.

Note that the equations [MATH]a=\sqrt{4-\sqrt{5-a}}[/MATH] and [MATH]b=\sqrt{4+\sqrt{5-b}}[/MATH] are closely related and both $a$ and $b$ do satisfy the equation

$\color{blue}{(x^2-4)^2 = 5-x}$, or $\color{blue}{x^4 - 8x^2 + x + 11 = 0}$

We are not making this equation up, nor we have the crystal ball to tell us this indeed is the quartic equation that $a$ and $b$ satisfy the equation. We actually worked it out for that quartic equation in the following manner:

From [MATH]a=\sqrt{4-\sqrt{5-a}}[/MATH], or more generally, [MATH]x=\sqrt{4-\sqrt{5-x}}[/MATH], we see that if we square both sides of the equation and then subtract both sides by 4 and again, square the result of it, we have:

[MATH]x^2=4-\sqrt{5-x}[/MATH]

[MATH]x^2-4=-\sqrt{5-x}[/MATH]

[MATH](x^2-4)^2=5-x[/MATH]

[MATH]\therefore x^2-4=\pm\sqrt{5-x}[/MATH] and

[MATH]x=\pm\sqrt{4\pm\sqrt{5-x}}[/MATH]

Taking only the positive roots, we have

[MATH]x=\sqrt{4\pm\sqrt{5-x}}[/MATH]

i.e.

[MATH]a=\sqrt{4-\sqrt{5-a}}[/MATH], and [MATH]b=\sqrt{4+\sqrt{5-b}}[/MATH]

And by the similar token,

Note that the equations [MATH]c=\sqrt{4-\sqrt{5+c}}[/MATH] and  [MATH]d=\sqrt{4+\sqrt{5+d}}[/MATH] are closely related and both $c$ and $d$ do satisfy the equation

$\color{red}{(x^2-4)^2 = 5+x}$, or $\color{red}{x^4 - 8x^2 - x + 11 = 0}$

since

From  [MATH]c=\sqrt{4-\sqrt{5+c}}[/MATH], or more generally, [MATH]x=\sqrt{4-\sqrt{5+x}}[/MATH], and repeat doing what we did for the above, we get

[MATH]x^2=4-\sqrt{5+x}[/MATH]

[MATH]x^2-4=-\sqrt{5+x}[/MATH]

[MATH](x^2-4)^2=5+x[/MATH]

[MATH]\therefore x^2-4=\pm\sqrt{5+x}[/MATH] and

[MATH]x=\pm\sqrt{4\pm\sqrt{5+x}}[/MATH]

Taking only the positive roots, we have

[MATH]x=\sqrt{4\pm\sqrt{5+x}}[/MATH]

i.e.

[MATH]c=\sqrt{4-\sqrt{5+c}}[/MATH], and [MATH]d=\sqrt{4+\sqrt{5+d}}[/MATH]

Now, how could this helps to compute the value of $abcd$???

Don't feel glum and just "sit" around...we need to keep a really open mind to see what sort of information could we glean from these two equations:

[MATH]\color{blue}{x^4 - 8x^2 + x + 11 = 0}[/MATH]

[MATH]\color{red}{x^4 - 8x^2 - x + 11 = 0}[/MATH]

We know $abcd$ is the product of 4 real numbers, and this should remind us of the roots and its polynomials, that is the cue we could get from this and if we think along with this line, we see that:

If $x$ satisfies [MATH]\color{red}{x^4 - 8x^2 - x + 11 = 0}[/MATH], then $-x$ satisfies [MATH]\color{blue}{x^4 - 8x^2 + x + 11 = 0}[/MATH].

So the roots of

[MATH]\color{blue}{x^4 - 8x^2 + x + 11 = 0}[/MATH] are $a,b,-c,-d.$

Thus $abcd=11$.