x^3 −x−c=0 let x=(√(t+p))+(√(t+q)) (t+p)(√(t+p))+(t+q)(√(t+q)) +3(t+p)(√(t+q))+3(t+q)(√(t+p)) −(√(t+p))−(√(t+q))−c=0 let (t+q)+3(t+p)−1=0 ⇒ 4t=1−(p+q) ⇒ 4t+4p=3p−q+1 4t+4q=1−(p−3q) (3p−q+1)^(3/2) +3(1−p+3q)(√(3p−q+1)) −4(√(3p−q+1))−8c=0 2x=(√(3p−q+1))+(√(3q−p+1)) let (√(3p−q+1))=−1 ⇒ q=3p ⇒ −1−3(1−p+3q)+4−8c=0 ⇒ p=−(c/3) 2x=−1+(√(1−((8c)/3))) x=−(1/2)+(√((1/4)−((2c)/3))) for c=(1/4) x=−(1/2)+(1/(2(√3)))


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Question Number 144768 by ajfour last updated on 04/Jul/21

$x^{3} - x - c = 0$ $l e t x = \sqrt{t + p} + \sqrt{t + q}$ $(t + p) \sqrt{t + p} + (t + q) \sqrt{t + q}$ $+ 3 (t + p) \sqrt{t + q} + 3 (t + q) \sqrt{t + p}$ $- \sqrt{t + p} - \sqrt{t + q} - c = 0$ $l e t (t + q) + 3 (t + p) - 1 = 0$ $\Rightarrow 4 t = 1 - (p + q)$ $\Rightarrow 4 t + 4 p = 3 p - q + 1$ $4 t + 4 q = 1 - (p - 3 q)$ ${(3 p - q + 1)}^{3 / 2} + 3 (1 - p + 3 q) \sqrt{3 p - q + 1}$ $- 4 \sqrt{3 p - q + 1} - 8 c = 0$ $2 x = \sqrt{3 p - q + 1} + \sqrt{3 q - p + 1}$ $l e t \sqrt{3 p - q + 1} = - 1$ $\Rightarrow q = 3 p$ $\Rightarrow - 1 - 3 (1 - p + 3 q) + 4 - 8 c = 0$ $\Rightarrow p = - \frac{c}{3}$ $2 x = - 1 + \sqrt{1 - \frac{8 c}{3}}$ $x = - \frac{1}{2} + \sqrt{\frac{1}{4} - \frac{2 c}{3}}$ $f o r c = \frac{1}{4}$ $x = - \frac{1}{2} + \frac{1}{2 \sqrt{3}}$
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