find-all-real-solutions-2-x-2-x-2-3-2x-4-1-i-what-if-x-is-permitted-to-be-complex-number-ii-what-if-1-1-2n-

Question Number 37382 by mondodotto@gmail.com last updated on 12/Jun/18

find all real solutions

{(2 - x^{2})}^{x^{2} - 3 \sqrt{2 x} + 4} = 1

i} what if x is permitted

to be complex number

ii} what if 1 = {(- 1)}^{2 n} ?

Commented by math khazana by abdo last updated on 12/Jun/18

(e) \Rightarrow 2 - x^{2} = 1 o r x^{2} - 3 \sqrt{2} x + 4 = 0 b u t

2 - x^{2} = 1 \Rightarrow x^{2} = 1 \Rightarrow x = \bar{+} 1

x^{2} - 3 \sqrt{2} x + 4 = 0 Δ = {(- 3 \sqrt{2})}^{2} - 4.4 = 18 - 16 = 2

x_{1} = \frac{3 \sqrt{2} + \sqrt{2}}{2} = 2 \sqrt{2} a n d x_{2} = \frac{3 \sqrt{2} - \sqrt{2}}{2} = \sqrt{2}

i f x i s a t i n t e r i o r o f \sqrt{\dots} c h a n g . \sqrt{x} = t g i v e

t^{4} - 3 \sqrt{2} t + 4 = 0 Δ = 18 - 16 = 2

t^{2} = 2 \sqrt{2} o r t^{2} = \sqrt{2} \Rightarrow t = \bar{+} \sqrt{2 \sqrt{2}} o r t = \bar{+} \sqrt{\sqrt{2}}

Commented by MJS last updated on 13/Jun/18

{it}^{'} s not t^{4} - 3 \sqrt{2} t^{2} + 4 = 0 but t^{4} - 3 \sqrt{2} t + 4 = 0

and {it}^{'} s not this easy to solve it (not sure if

{it}^{'} s possible to find exact solutions)

Commented by math khazana by abdo last updated on 13/Jun/18

n o s i r M j s t h e e q u a t i o n i s x^{2} - 3 \sqrt{2 x} + 4 = 0

s o i f w e s e t x = t^{2} w e g e t t^{4} - 3 \sqrt{2} t + 4 = 0

t h e r e i s n o e r r o r s i r (i h a v e c o n s i d e r e d x a t

t h e i n t e r i o r o f \sqrt{\dots})

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jun/18

x^{2} - 3 \sqrt{2} x + 4 = 0

s i n c e {(2 - x^{2})}^{x^{2} - 3 \sqrt{2} x + 4} = 1 = {(2 - x^{2})}^{0}

x^{2} - 2 . x . \frac{3 \sqrt{2}}{2} + \frac{9}{2} + 4 - \frac{9}{2} = 0

{(x - \frac{3}{\sqrt{2}})}^{2} - \frac{1}{2} = 0 i n p l a c e o f \frac{9}{2} i p u t \frac{9}{4} c o r r e c t d

x - \frac{3}{\sqrt{2}} = \pm \frac{1}{\sqrt{2}}

x = \frac{4}{\sqrt{2}} a n d \frac{2}{\sqrt{2}} x = 2 \sqrt{2} a n d \sqrt{2}

Commented by mondodotto@gmail.com last updated on 12/Jun/18

does this satisfy the whole equation ? ? remember

the first part we only need real number

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Jun/18

2 - x^{2} = 1

x^{2} = 1

x = \pm 1

i t h o u g h t t h i s p o r t i o n i s w e l u n d e r s t o o d

Commented by MJS last updated on 12/Jun/18

solutions of x^{2} - 3 \sqrt{2} x + 4 = 0 are real

Commented by prakash jain last updated on 12/Jun/18

question has \sqrt{2 x} not \sqrt{2} x

Answered by MJS last updated on 12/Jun/18

x \in R

(x^{2} - 3 \sqrt{2} x + 4) \ln (2 - x^{2}) = 0

x^{2} - 3 \sqrt{2} x + 4 = 0

x = \frac{3 \sqrt{2}}{2} \pm \sqrt{{(\frac{3 \sqrt{2}}{2})}^{2} - 4} = \frac{3 \sqrt{2}}{2} \pm \frac{\sqrt{2}}{2}

x_{1} = \sqrt{2}

[this leads to 0^{0} which is not defined but

I think \lim_{x \to \sqrt{2}} {(2 - x^{2})}^{x^{2} - 3 \sqrt{2} x + 4} = 1 so we can

count it]

x_{2} = 2 \sqrt{2}

2 - x^{2} = 1

x^{2} = 1

x_{3} = - 1

x_{4} = 1

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