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- #1

- Thread starter jacks
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- Thread starter
- #1

- Nov 29, 2013

- 172

Hello.If $a,b,c,d$ are distinct real no. such that

$a=\sqrt{4+\sqrt{5+a}}\;,b=\sqrt{4-\sqrt{5+b}}\;,c=\sqrt{4+\sqrt{5-c}}\;,d=\sqrt{4-\sqrt{5-d}}$. Then $abcd=$

[tex]b^4-8b^2-b+11=0[/tex]

[tex]c^4-8c^2+c+11=0[/tex]

[tex]d^4-8d^2+d+11=0[/tex]

Common roots "a" and "b":

[tex]r_1=-2.3710[/tex]

[tex]r_2=-1.4551[/tex]

[tex]r_3=1.2266[/tex]

[tex]r_4=2.5994[/tex]

Common roots "c" and "d":

[tex]s_1=2.3710[/tex]

[tex]s_2=1.4551[/tex]

[tex]s_3=-1.2266[/tex]

[tex]s_4=-2.5994[/tex]

There are several solutions product of roots.

A curiosity, jacks. Do you participate in a forum on Spanish?

Regards.

- Moderator
- #3

- Feb 7, 2012

- 2,799

In order to get a unique solution, I believe the question should sayIf $a,b,c,d$ are distinct real no. such that

$a=\sqrt{4+\sqrt{5+a}}\;,b=\sqrt{4-\sqrt{5+b}}\;,c=\sqrt{4+\sqrt{5-c}}\;,d=\sqrt{4-\sqrt{5-d}}$. Then $abcd=$

If $a,b,c,d$ are distinct *positive* real nos. such that

$a=\sqrt{4+\sqrt{5+a}}\;,b=\sqrt{4-\sqrt{5+b}}\;,c=\sqrt{4+\sqrt{5-c}}\;,d=\sqrt{4-\sqrt{5-d}}$. Then $abcd=$ ?

$a=\sqrt{4+\sqrt{5+a}}\;,b=\sqrt{4-\sqrt{5+b}}\;,c=\sqrt{4+\sqrt{5-c}}\;,d=\sqrt{4-\sqrt{5-d}}$. Then $abcd=$ ?

Now it is clear that $x$ is a solution of $x^4 - 8x^2 + x + 11 = 0$ if and only if $-x$ is a solution of $x^4 - 8x^2 - x + 11 = 0$. Also, the sum of the roots of $x^4 - 8x^2 + x + 11$ is $0$, and their product is $11$. Therefore, given that the roots are real, two of them must be positive and two negative. It follows that the numbers $a,b,c,d$ must be the absolute values of the roots of $x^4 - 8x^2 + x + 11$, and therefore their product is $11$.