solve for:a,b,c,d∈R a^2 =b+(√c) b^2 =c+(√d) c^2 =d+(√a) d^2 =a+(√b)


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 54786 by behi83417@gmail.com last updated on 10/Feb/19

$s o l v e f o r : a, b, c, d \in R$ $a^{2} = b + \sqrt{c}$ $b^{2} = c + \sqrt{d}$ $c^{2} = d + \sqrt{a}$ $d^{2} = a + \sqrt{b}$
Commented by mr W last updated on 11/Feb/19

$a = b = c = d = 0 o r {(\frac{\sqrt[3]{9 + \sqrt{69}} + \sqrt[3]{9 - \sqrt{69}}}{\sqrt[3]{18}})}^{2} \approx 1.7548 ?$
Commented by behi83417@gmail.com last updated on 11/Feb/19

$r i g h t a n s w e r s i r . p l e a s e p o s t y o u r$ $s o l u t i o n . t h a n k y o u v e r y m u c h .$
Commented by mr W last updated on 11/Feb/19

$d u e t o s y m m e t r y : a = b = c = d = t^{2}$ $t^{4} = t^{2} + t$ $t (t^{3} - t - 1) = 0$ $\Rightarrow t = 0$ $\Rightarrow t^{3} - t - 1 = 0$ $l e t t = u + v$ ${(u + v)}^{3} - 3 u v (u + v) - (u^{3} + v^{3}) = 0$ $u^{3} v^{3} = \frac{1}{27}$ $u^{3} + v^{3} = 1$ $x^{2} - x + \frac{1}{27} = 0$ $u^{3} (v^{3}) = \frac{1}{2} (1 \pm \sqrt{1 - \frac{4}{27}}) = \frac{1}{18} (9 \pm \sqrt{69})$ $u = \frac{\sqrt[3]{9 + \sqrt{69}}}{\sqrt[3]{18}}$ $v = \frac{\sqrt[3]{9 - \sqrt{69}}}{\sqrt[3]{18}}$ $\Rightarrow t = \frac{\sqrt[3]{9 + \sqrt{69}} + \sqrt[3]{9 - \sqrt{69}}}{\sqrt[3]{18}}$ $a = b = c = d = t^{2} = 0 o r {(\frac{\sqrt[3]{9 + \sqrt{69}} + \sqrt[3]{9 - \sqrt{69}}}{\sqrt[3]{18}})}^{2}$
Commented by behi83417@gmail.com last updated on 12/Feb/19

$t h a n k s a g a i n s i r . i t i s p e r f e c t .$ $i s t h e r e a n y s u l o t i o n w i t h o u t u s i n g$ $s y m m e t r y ?$
Commented by mr W last updated on 12/Feb/19

$i a m n o t s u r e i f t h e r e a r e o t h e r s o l u t i o n s .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum