Solve-in-R-x-2-4x-40x-2-32x-16-

Question Number 156809 by MathSh last updated on 15/Oct/21

Solve in R

x^{2} + 4 x = \sqrt{{40 x}^{2} + 32 x - 16}

Commented by MathSh last updated on 15/Oct/21

Yes dear S er

Answered by Rasheed.Sindhi last updated on 15/Oct/21

x (x + 4) = \sqrt{{40 x}^{2} + 32 x + 16}

x^{2} {(x + 4)}^{2} = {40 x}^{2} + 32 x + 16

x^{4} + {8 x}^{3} + {16 x}^{2} - {40 x}^{2} - 32 x - 16 = 0

x^{4} + {8 x}^{3} - {24 x}^{2} - 32 x - 16 = 0

x = 3.1850, - 10.0814 (U s i n g c a l c u l a t o r)

Y o u c a n s e e e x a c t r e s u l t s i n a n

a d v a n c e d c a l c u l a t o r b y y o u r s e l f .

{T h e y}^{'} r e t o o c o m p l i c a t e d!

Answered by MJS_new last updated on 16/Oct/21

x^{2} + 4 x = \sqrt{40 x^{2} + 32 x - 16}

squaring & transforming

x^{4} + 8 x^{3} - 24 x^{2} - 32 x + 16 = 0

(x^{2} + 4 (1 - \sqrt{2}) x - 4) (x^{2} + 4 (1 + \sqrt{2}) x - 4) = 0

of the 4 solutions 3 fit the given equation

x_{1} = - 2 + 2 \sqrt{2} + 2 \sqrt{4 - 2 \sqrt{2}}

x_{2, 3} = - 2 - 2 \sqrt{2} \pm 2 \sqrt{4 + 2 \sqrt{2}}

Commented by MJS_new last updated on 16/Oct/21

yes . if the equation for a has only 1 real

solution (Cardano) or if it has 3 real solutions

of trigonometric shape we cannot use them .

we must hope for easy solutions \in Q to be

found within the factors of \pm n

Commented by MathSh last updated on 16/Oct/21

Very nice dear S er thanks

Commented by Rasheed.Sindhi last updated on 16/Oct/21

T h a n k s f o r p r e c i o u s k n o w l e d g e!

Commented by MJS_new last updated on 16/Oct/21

{you}^{'} re welcome

Commented by Rasheed.Sindhi last updated on 16/Oct/21

S i r, h o w d i d y o u f a c t o r t h e e q u a t i o n ?

Commented by MJS_new last updated on 16/Oct/21

the method I use :

x^{4} + {a x}^{3} + {b x}^{2} + c x + d = 0

let x = t - \frac{a}{4} to get

t^{4} + {p t}^{2} + q t + r = 0

we want to find α, β, γ to get

(t^{2} - α t - β) (t^{2} + α t - γ) = 0

t^{4} - (α^{2} + β + γ) t^{2} + α (γ - β) t + β γ = 0

\Rightarrow {\begin{cases} p = - (α^{2} + β + γ) \\ q = α (γ - β) \\ r = β γ \end{cases}

solve the 1^{st} & 2^{nd} for β & γ

inserting in the 3^{rd} & transforming we get

α^{6} + j α^{4} + k α^{2} + l = 0

let α = \sqrt{a - \frac{j}{3}} to get

a^{3} + m a + n = 0

if this has an “ easy to {handle}^{″} solution we

succeed, otherwise the given equation has

no useable exact solution .

Commented by MJS_new last updated on 16/Oct/21

x^{2} + 4 x = \sqrt{40 x^{2} + 32 x + 16} has no useable exact solution

Commented by Rasheed.Sindhi last updated on 16/Oct/21

T h a n k s a l o t s i r! I u s e d a c a l c u l a t o r

a n d i t g a v e m e t e r r i b l e a n s w e r s

i n e x a c t f o r m!