Solve for x∈R x^4 +4x−3=0


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Question Number 203002 by mr W last updated on 07/Jan/24

$S o l v e f o r x \in R$ $x^{4} + 4 x - 3 = 0$
	Answered by ajfour last updated on 07/Jan/24

	$p^{2} (p^{4} + 12) = 16$ $\Rightarrow {(p^{2} / 2)}^{3} + 3 {(p^{2} / 2)}^{2} = 2$ $\frac{p^{2}}{2} = \sqrt[3]{\sqrt{2} + 1} - \sqrt[3]{\sqrt{2} - 1}$ $n o w x i s e a s y .$
	Commented by ajfour last updated on 07/Jan/24

	Answered by mr W last updated on 07/Jan/24

	$s o l v e x^{4} + 4 x - 3 = 0$ ${l e t}^{'} s g e n e r a l l y l o o k a t t h e e q u a t i o n$ $x^{4} + p x + q = 0 .$ $s a y f (x) = x^{4} + p x + q$ $\lim_{x \to - \infty} f (x) = + \infty$ $\lim_{x \to + \infty} f (x) = + \infty$ $f^{'} (x) = 4 x^{3} + p = 0 \Rightarrow x = - \sqrt[3]{\frac{p}{4}}$ $t h a t m e a n s f (x) h a s m i n i m u m a t$ $x_{m} = - {(\frac{p}{4})}^{\frac{1}{3}} .$ $f {(x)}_{m i n} = {[- {(\frac{p}{4})}^{\frac{1}{3}}]}^{4} + p [- {(\frac{p}{4})}^{\frac{1}{3}}] + q = - 3 {(\frac{p}{4})}^{\frac{4}{3}} + q$ $t h e r e a r e t h r e e c a s e s, s e e d i a g r a m$ $b e l o w .$ $c a s e 1 :$ $f {(x)}_{m i n} = - 3 {(\frac{p}{4})}^{\frac{4}{3}} + q > 0$ $i . e . {(\frac{p}{4})}^{4} - {(\frac{q}{3})}^{3} < 0$ $\Rightarrow f (x) = 0 h a s n o r e a l r o o t .$ $c a s e 2 :$ $f {(x)}_{m i n} = - 3 {(\frac{p}{4})}^{\frac{4}{3}} + q = 0$ $i . e . {(\frac{p}{4})}^{4} - {(\frac{q}{3})}^{3} = 0$ $\Rightarrow f (x) = 0 h a s a d o u b l e r e a l r o o t$ $x_{1, 2} = x_{m} = - {(\frac{p}{4})}^{\frac{1}{3}}$ $c a s e 3 :$ $f {(x)}_{m i n} = - 3 {(\frac{p}{4})}^{\frac{4}{3}} + q = p < 0$ $i . e . {(\frac{p}{4})}^{4} - {(\frac{q}{3})}^{3} > 0$ $\Rightarrow f (x) = 0 h a s t w o r e a l r o o t s .$
	Commented by mr W last updated on 07/Jan/24

	Commented by mr W last updated on 07/Jan/24

	$x^{4} + p x + q = 0$ $(x^{2} + a x + b) (x^{2} + c x + d) = x^{4} + p x + q$ $x^{4} + (a + c) x^{3} + (b + d + a c) x^{2} + (b c + a d) x + b d = x^{4} + p x + q$ $a + c = 0 \Rightarrow c = - a$ $b + d + a c = 0 \Rightarrow d = a^{2} - b = \frac{a^{2}}{2} + \frac{p}{2 a}$ $b c + a d = p \Rightarrow a^{3} - 2 b a = p \Rightarrow b = \frac{a^{3} - p}{2 a} = \frac{a^{2}}{2} - \frac{p}{2 a}$ $b d = q \Rightarrow b (a^{2} - b) = q$ $\Rightarrow (\frac{a^{3} - p}{2 a}) (a^{2} - \frac{a^{3} - p}{2 a}) = q$ $\Rightarrow a^{6} - 4 {q a}^{2} - p^{2} = 0$ $Δ = {(- \frac{4 q}{3})}^{3} + {(- \frac{p^{2}}{2})}^{2} = 64 [{(\frac{p}{4})}^{4} - {(\frac{q}{3})}^{3}]$ $i f {(\frac{p}{4})}^{4} - {(\frac{q}{3})}^{3} ⩾ 0, Δ ⩾ 0$ $\Rightarrow a^{2} = \sqrt[3]{8 \sqrt{{(\frac{p}{4})}^{4} - {(\frac{q}{3})}^{3}} + \frac{p^{2}}{2}} - \sqrt[3]{8 \sqrt{{(\frac{p}{4})}^{4} - {(\frac{q}{3})}^{3}} - \frac{p^{2}}{2}}$ $\Rightarrow a = \pm \sqrt{\sqrt[3]{8 \sqrt{{(\frac{p}{4})}^{4} - {(\frac{q}{3})}^{3}} + \frac{p^{2}}{2}} - \sqrt[3]{8 \sqrt{{(\frac{p}{4})}^{4} - {(\frac{q}{3})}^{3}} - \frac{p^{2}}{2}}}$ $f o r r e a l r o o t s w e t a k e t h e s a m e s i g n$ $f o r a a s t h e s i g n o f p .$ $e x a m p l e w i t h p = 4, q = - 3$ $a = \sqrt{2 (\sqrt[3]{\sqrt{2} + 1} - \sqrt[3]{\sqrt{2} - 1})}$ $x_{1, 2} = \frac{- a \pm \sqrt{a^{2} - 4 b}}{2}$ $= \frac{1}{2} (- a \pm \sqrt{\frac{2 p}{a} - a^{2}})$ $\approx - 1.7844 \lor 0.6925$
	Commented by ajfour last updated on 07/Jan/24
	yes this is the simpler in fact. Thank you sir. but b=0 in my formula for x would be equivalently simple too.
	Commented by mr W last updated on 07/Jan/24

	$y o u a r e r i g h t s i r! t h a n k s!$
	Answered by Frix last updated on 07/Jan/24

	$Just the same but another point of view .$ $x^{4} + p x + q = 0$ $p = - 4 α (α^{2} + β)$ $q = (α^{2} - β) (3 α^{2} + β)$ $x^{4} - 4 α (α^{2} + β) x + (α^{2} - β) (3 α^{2} + β) = 0$ $(x^{2} - 2 α x + α^{2} - β) (x^{2} + 2 α x + 3 α^{2} - β) = 0$ $x_{1, 2} = α \pm \sqrt{β}$ $x_{2, 3} = - α \pm \sqrt{- (2 α^{2} + β)}$ $We end up with$ $α^{6} - \frac{q α^{2}}{4} - \frac{p^{2}}{64} = 0 \land β = - α^{2} - \frac{p}{4 α}$
	Commented by mr W last updated on 07/Jan/24

	$t h a n k s s i r!$
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