Solve-for-x-R-x-4-4x-3-0-

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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