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Tinku Tara June 4, 2023 Algebra 0 Comments

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Question Number 82191 by jagoll last updated on 19/Feb/20

find the solution (√(x^2 −3x−4 )) > x−2

f i n d t h e s o l u t i o n

\sqrt{x^{2} - 3 x - 4} > x - 2

Commented by arkanmath7@gmail.com last updated on 19/Feb/20

x^2 −3x−4 > x^2 −4x+4 −3x−4 > −4x+4 x > 8 S.S. = {x:x>8} Or S.S.={(8,∞)} so easy

x^{2} - 3 x - 4 > x^{2} - 4 x + 4

- 3 x - 4 > - 4 x + 4

x > 8

S . S . = {x : x > 8}

O r S . S . = {(8, \infty)}

s o e a s y

Commented by jagoll last updated on 19/Feb/20

tes for x = −1 (√(1+3−4)) > −3 0 > −3 is correct sir

t e s f o r x = - 1

\sqrt{1 + 3 - 4} > - 3

0 > - 3 i s c o r r e c t s i r

Commented by jagoll last updated on 19/Feb/20

x = −1 ∉ (8,∞) but x=−1 is solution

x = - 1 \notin (8, \infty) b u t x = - 1 i s s o l u t i o n

Commented by mr W last updated on 19/Feb/20

(√((x−4)(x+1)))>x−2 solution: x≤−1 ∨ x>8

\sqrt{(x - 4) (x + 1)} > x - 2

s o l u t i o n :

x ⩽ - 1 \lor x > 8

Commented by arkanmath7@gmail.com last updated on 19/Feb/20

I think that′s true if you said (√(x^2 −3x−4)) ≥0 not when (√(x^2 −3x−4)) >x−2 put x=−1 in the 2nd line of the solution you will get a rong statement

I t h i n k {t h a t}^{'} s t r u e i f y o u s a i d \sqrt{x^{2} - 3 x - 4} ⩾ 0

n o t w h e n \sqrt{x^{2} - 3 x - 4} > x - 2

p u t x = - 1 i n t h e 2 n d l i n e o f t h e s o l u t i o n

y o u w i l l g e t a r o n g s t a t e m e n t

Commented by Kunal12588 last updated on 19/Feb/20

x > 8 and (√((x+1)(x−4)))>(x−2) ⇒(((x+1)(x−4))/((x−2)^2 ))>0 ⇒x<−1

x > 8

a n d \sqrt{(x + 1) (x - 4)} > (x - 2)

\Rightarrow \frac{(x + 1) (x - 4)}{{(x - 2)}^{2}} > 0

\Rightarrow x < - 1

Commented by mr W last updated on 19/Feb/20

why use if (√(x^2 −3x−4))≥0 ? it is always true that (√(x^2 −3x−4))≥0.

w h y u s e i f \sqrt{x^{2} - 3 x - 4} ⩾ 0 ? i t i s a l w a y s

t r u e t h a t \sqrt{x^{2} - 3 x - 4} ⩾ 0 .

Commented by jagoll last updated on 19/Feb/20

so what is the answer conclusion? x≤−1 ∨x> 8 or x > 8?

s o w h a t i s t h e a n s w e r c o n c l u s i o n ?

x ⩽ - 1 \lor x > 8

o r x > 8 ?

Commented by mr W last updated on 19/Feb/20

when x≤−1, x−2≤−3, but (√(x^2 −3x−4))≥0, therefore (√(x^2 −3x−4))>x−2 is true. and x=−1 is also included.

w h e n x ⩽ - 1, x - 2 ⩽ - 3, b u t

\sqrt{x^{2} - 3 x - 4} ⩾ 0, t h e r e f o r e

\sqrt{x^{2} - 3 x - 4} > x - 2 i s t r u e . a n d x = - 1

i s a l s o i n c l u d e d .

Commented by mr W last updated on 19/Feb/20

the result is x≤−1 ∨ x> 8 that means x∈ (−∞,−1] or x∈(8,+∞)

t h e r e s u l t i s

x ⩽ - 1 \lor x > 8

t h a t m e a n s x \in (- \infty, - 1] o r x \in (8, + \infty)

Commented by jagoll last updated on 19/Feb/20

yes sir. i agree. but my teacher blamed my answer

y e s s i r . i a g r e e . b u t m y t e a c h e r

b l a m e d m y a n s w e r

Commented by arkanmath7@gmail.com last updated on 19/Feb/20

confused quest i think you are the truest

c o n f u s e d q u e s t

i t h i n k y o u a r e t h e t r u e s t

Commented by jagoll last updated on 19/Feb/20

haha...if the most correct , surely God sir

h a h a \dots i f t h e m o s t c o r r e c t,

s u r e l y G o d s i r

Commented by mr W last updated on 19/Feb/20

be very careful with squaring by inequlity! it may change the validity range of the original inequality!

b e v e r y c a r e f u l w i t h s q u a r i n g b y

i n e q u l i t y! i t m a y c h a n g e t h e v a l i d i t y

r a n g e o f t h e o r i g i n a l i n e q u a l i t y!

Commented by jagoll last updated on 19/Feb/20

please sir post your step solution

p l e a s e s i r p o s t y o u r s t e p s o l u t i o n

Commented by jagoll last updated on 19/Feb/20

Commented by jagoll last updated on 19/Feb/20

it sir by graphic

i t s i r b y g r a p h i c

Commented by mathmax by abdo last updated on 19/Feb/20

the inequation is defined for x≥2 and x^2 −3x−4≥0 Δ=(−3)^2 −4(−4)=9+16=25 ⇒x_1 =((3+5)/2)=4 and x_2 =((3−5)/2)=−1 x^2 −3x−4≥0 ⇒x∈]−∞,−1[∪]4,+∞[ ⇒ D_(in) =[2,+∞[ (in)⇒x^2 −3x−4>x^2 −4x +4 ⇒x>8 ⇒S =[8,+∞[

t h e i n e q u a t i o n i s d e f i n e d f o r x ⩾ 2 a n d x^{2} - 3 x - 4 ⩾ 0

Δ = {(- 3)}^{2} - 4 (- 4) = 9 + 16 = 25 \Rightarrow x_{1} = \frac{3 + 5}{2} = 4 a n d x_{2} = \frac{3 - 5}{2} = - 1

x^{2} - 3 x - 4 ⩾ 0 \Rightarrow x \in] - \infty, - 1 [\cup] 4, + \infty [\Rightarrow D_{i n} = [2, + \infty [

(i n) \Rightarrow x^{2} - 3 x - 4 > x^{2} - 4 x + 4 \Rightarrow x > 8 \Rightarrow S = [8, + \infty [

Commented by jagoll last updated on 19/Feb/20

solution x≤−1 ∨ x>8

s o l u t i o n x ⩽ - 1 \lor x > 8

Answered by mr W last updated on 19/Feb/20

such that (√(x^2 −3x−4)) is defined, x^2 −3x−4=(x+1)(x−4)≥0 ⇒x≤−1 or x≥4 case 1: x≤−1 with x≤−1, we have x−2≤−3<0 but (√(x^2 −3x−4))≥0, so (√(x^2 −3x−4 )) > x−2 is true. ⇒x≤−1 is solution. case 2: x≥4 with x≥4, x−2≥6>0 x^2 −3x−4 >(x−2)^2 x^2 −3x−4 >x^2 −4x+4 x>8 ≥4 ⇒x>8 is solution. (we can use squaring here because we have ensured at first that x−2>0) summary of all solutions: x≤−1 x>8

s u c h t h a t \sqrt{x^{2} - 3 x - 4} i s d e f i n e d,

x^{2} - 3 x - 4 = (x + 1) (x - 4) ⩾ 0

\Rightarrow x ⩽ - 1 o r x ⩾ 4

c a s e 1 : x ⩽ - 1

w i t h x ⩽ - 1, w e h a v e x - 2 ⩽ - 3 < 0

b u t \sqrt{x^{2} - 3 x - 4} ⩾ 0, s o

\sqrt{x^{2} - 3 x - 4} > x - 2 i s t r u e .

\Rightarrow x ⩽ - 1 i s s o l u t i o n .

c a s e 2 : x ⩾ 4

w i t h x ⩾ 4, x - 2 ⩾ 6 > 0

x^{2} - 3 x - 4 > {(x - 2)}^{2}

x^{2} - 3 x - 4 > x^{2} - 4 x + 4

x > 8 ⩾ 4

\Rightarrow x > 8 i s s o l u t i o n .

(w e c a n u s e s q u a r i n g h e r e b e c a u s e w e

h a v e e n s u r e d a t f i r s t t h a t x - 2 > 0)

s u m m a r y o f a l l s o l u t i o n s :

x ⩽ - 1

x > 8

Commented by jagoll last updated on 19/Feb/20

thank you sir

t h a n k y o u s i r

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