Question-212518

Question Number 212518 by efronzo1 last updated on 16/Oct/24

\underset{⏟}{⇃} \underset{―}{}

Answered by golsendro last updated on 16/Oct/24

\underset{―}{\underset{⏟}{\pm}} = {(x - a)}^{2} + a

f^{- 1} (x) = \sqrt{x - a} + a

\Leftrightarrow {(x - a)}^{2} = \sqrt{x - a}

\sqrt{x - a} (\sqrt{{(x - a)}^{3}} - 1) = 0

x = a or x = a + 1

so if x = 5049 = a then the other solution

Answered by mr W last updated on 16/Oct/24

x^{2} - 2 a x + a (a + 1) = x

x^{2} - (2 a + 1) x + a (a + 1) = 0

s a y t h e o t h e r s o l u t i o n i s β .

5049 + β = 2 a + 1

5049 β = a (a + 1)

4 \times 5049 β = {(5049 + β)}^{2} - 1

{(β - 5049)}^{2} - 1 = 0

\Rightarrow β = 5049 \pm 1

i . e . t h e o t h e r s o l u t i o n i s 5048 o r 5050 .

w i t h a = 5048 o r 5049 .

Commented by mr W last updated on 16/Oct/24

Commented by mr W last updated on 16/Oct/24

Commented by golsendro last updated on 16/Oct/24

f : [a, \infty) \to [a, \infty) sir

it follow that β = 5059 - 1 not

solution

Commented by mr W last updated on 16/Oct/24

w i t h a = 5048 t h e r e a r e t w o s o l u t i o n s :

x = 5048, 5049

w i t h a = 5049 t h e r e a r e t w o s o l u t i o n s :

x = 5049, 5050

w h a t i s w r o n g ?

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