MHB Product of roots abcd in 4th degree equation

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The discussion revolves around finding the product of the roots \( abcd \) of the equation defined by \( a, b, c, d \) as distinct real numbers satisfying specific square root equations. It is suggested that the problem should specify that \( a, b, c, d \) are distinct positive real numbers for a unique solution. The roots of the polynomial \( x^4 - 8x^2 + x + 11 = 0 \) are analyzed, revealing that the sum of the roots is zero and the product is 11. Consequently, it is concluded that the product \( abcd \) equals 11. The final result confirms that the product of the roots is indeed 11.
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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 $abcd=$
 
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jacks said:
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 $abcd=$

Hello.

a^4-8a^2-a+11=0

b^4-8b^2-b+11=0

c^4-8c^2+c+11=0

d^4-8d^2+d+11=0

Common roots "a" and "b":

r_1=-2.3710

r_2=-1.4551

r_3=1.2266

r_4=2.5994

Common roots "c" and "d":

s_1=2.3710

s_2=1.4551

s_3=-1.2266

s_4=-2.5994

There are several solutions product of roots.

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

Regards.
 
jacks said:
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 $abcd=$
In order to get a unique solution, I believe the question should say
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 $abcd=$ ?​
[sp]Then $a,b,c,d$ all satisfy the equation $x = \sqrt{4 \pm\sqrt{5\pm x}}$. Square both sides to get $x^2 - 4 = \pm\sqrt{5\pm x}$. Square both sides again, getting $(x^2-4)^2 = 5 \pm x$, or $x^4 - 8x^2 \pm x + 11 = 0$.

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 $11$.[/sp]
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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