Solve-2x-5-x-1-gt-8-

Question Number 21354 by Tinkutara last updated on 21/Sep/17

Solve : \sqrt{2 x + 5} + \sqrt{x - 1} > 8

Commented by Tinkutara last updated on 22/Sep/17

W h e r e i s m y m i s t a k e ?

\sqrt{2 x + 5} + \sqrt{x - 1} > 8

2 x + 5 + x - 1 + 2 \sqrt{(2 x + 5) (x - 1)} > 64

2 \sqrt{(2 x + 5) (x - 1)} > 60 - 3 x

4 (2 x + 5) (x - 1) > 9 {(x - 20)}^{2}

8 x^{2} + 12 x - 20 > 9 x^{2} - 360 x + 3600

x^{2} - 372 x + 3620 < 0

(x - 362) (x - 10) < 0

x \in (10, 362)

W h y n o t I a m g e t t i n g x \in (10, \infty) ?

Commented by alex041103 last updated on 22/Sep/17

O k . F i r s t l e t s c o n s i d e r t h e f o l o w i n g

L e t

x < y

i f w e s q u a r e b o t h s i d e s w e g e t

x^{2} < y^{2}, b u t

x < y \to - x > - y \to {(- x)}^{2} > {(- y)}^{2}

o r x^{2} > y^{2} \to c o n t r a d i c t i o n .

B e f o r e w e e x p l o r e w h y t h i s i s a p r o b l e m

{l e t}^{'} s c o n s i d e r a n e a s i e r c a s e :

- x > 2 \Rightarrow x < - 2 (n o t >)

W e k n o w t h a t w e r e v e r s e t h e s i g n s .

S q u a r i n g b o t h s i d e s o f t h e i n e q u a l i t y

i s s i m i l a r t o m u l t i p l y i n g b o t h s i d e s

b y - 1 (o r t a k i n g t h e r e c i p r i c a l o f b o t h

s i d e s) .

S o {l e t}^{'} s c o n s i d e r t h e g e n e r a l c a s e :

i f g (x) ≷ h (x) t h e n f (g (x)) ≷ o r ≶ f (h (x)

f o r s o m e f u n c t i o n f (x) \neq c o n s t .

L e t s = s g n (\frac{d}{d x} f (x)) = c o n s t . f o r s o m e

x \in I w h e r e [g (x), h (x)] \in I (s u p p o s e t h e r e

i s s u c h I), t h e n

{\begin{cases} f (g (x)) ≷ f (h (x)), s = 1 \\ f (g (x)) ≶ f (h (x)), s = - 1 \end{cases}

I n c o n c l u s i o n w e d o r e v e r s e t h e i n e q u a l i t y

g ≷ h i f \frac{d}{d x} f (x) < 0 f o r x \in [g, h]

a n d w e {d o n}^{'} t i f \frac{d}{d x} f (x) > 0 f o r x \in [g, h]

I n o u r c a s e o f f (x) = x^{2}

s g n (\frac{d}{d x} f (x)) = s g n (x)

S o i n o r d e r t o b e e v e n a b l e t o s q u a r e

b o t h s i d e s t h e s i g n o f b o t h s i d e s m u s t

b e t h e s a m e .

I n t h e c a s e o f

\sqrt{2 x + 5} + \sqrt{x - 1} > 8

f o r x \in [1, \infty], s g n (\sqrt{2 x + 5} + \sqrt{x - 1}) = s g n (8) = 1

\Rightarrow T h e r e i s n o p r o b l e m i n s q u a r i n g

a n d w e {d o n}^{'} t r e v e r s e t h e i n e q u a l i t y

B u t f o r 2 \sqrt{(2 x + 5) (x - 1)} > 60 - 3 x

s g n (L H S) = 1 a n d s g n (60 - 3 x) = 1 o r - 1

W h e n x \in (20, \infty), s g n (60 - 3 x) = - 1

a n d t h e n 0 > 60 - 3 x a n d 2 \sqrt{(2 x + 5) (x - 1)} > 0

\Rightarrow x \in (20, \infty) t h e n 2 \sqrt{(2 x + 5) (x - 1)} > 60 - 3 x

W h e n x \in (1, 20) (s t a r t i n g f r o m o n e

i n o r d e r f o r t h e s q u a r e r o o t t o b e r e a l)

s g n (60 - 3 x) = 1

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

w e s e e t h a t f o r x \in (1, 20), x \in (10, 362)

o r x \in [10, 20] (t h e i n e q . i s t r u e f o r x = 20, 10)

\Rightarrow A n s . x \in [10, \infty)

Commented by Tinkutara last updated on 22/Sep/17

Thank you very much Sir!

Commented by alex041103 last updated on 22/Sep/17

Y e s . S o r r y a b o u t t h a t . {T h a t}^{'} s b e c a u s d

\sqrt{2 x + 5} + \sqrt{x - 1} = 8 w h e n x = 10