find the solution (√(x^2 −3x−4 ))


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

$f i n d t h e s o l u t i o n$ $\sqrt{x^{2} - 3 x - 4} > x - 2$
Commented byarkanmath7@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 byjagoll last updated on 19/Feb/20

$t e s f o r x = - 1$ $\sqrt{1 + 3 - 4} > - 3$ $0 > - 3 i s c o r r e c t s i r$
Commented byjagoll last updated on 19/Feb/20

$x = - 1 \notin (8, \infty) b u t x = - 1 i s s o l u t i o n$
Commented bymr W last updated on 19/Feb/20

$\sqrt{(x - 4) (x + 1)} > x - 2$ $s o l u t i o n :$ $x ⩽ - 1 \lor x > 8$
Commented byarkanmath7@gmail.com last updated on 19/Feb/20

$I t h i n k {t h a t}^{'} s t r u e i f y o u s a i d \sqrt{x^{2} - 3 x - 4} ⩾ 0$ $n o t w h e n \sqrt{x^{2} - 3 x - 4} > x - 2$ $p u t x = - 1 i n t h e 2 n d l i n e o f t h e s o l u t i o n$ $y o u w i l l g e t a r o n g s t a t e m e n t$
Commented byKunal12588 last updated on 19/Feb/20

$x > 8$ $a n d \sqrt{(x + 1) (x - 4)} > (x - 2)$ $\Rightarrow \frac{(x + 1) (x - 4)}{{(x - 2)}^{2}} > 0$ $\Rightarrow x < - 1$
Commented bymr W last updated on 19/Feb/20

$w h y u s e i f \sqrt{x^{2} - 3 x - 4} ⩾ 0 ? i t i s a l w a y s$ $t r u e t h a t \sqrt{x^{2} - 3 x - 4} ⩾ 0 .$
Commented byjagoll last updated on 19/Feb/20

$s o w h a t i s t h e a n s w e r c o n c l u s i o n ?$ $x ⩽ - 1 \lor x > 8$ $o r x > 8 ?$
Commented bymr W last updated on 19/Feb/20

$w h e n x ⩽ - 1, x - 2 ⩽ - 3, b u t$ $\sqrt{x^{2} - 3 x - 4} ⩾ 0, t h e r e f o r e$ $\sqrt{x^{2} - 3 x - 4} > x - 2 i s t r u e . a n d x = - 1$ $i s a l s o i n c l u d e d .$
Commented bymr W last updated on 19/Feb/20

$t h e r e s u l t i s$ $x ⩽ - 1 \lor x > 8$ $t h a t m e a n s x \in (- \infty, - 1] o r x \in (8, + \infty)$
Commented byjagoll last updated on 19/Feb/20

$y e s s i r . i a g r e e . b u t m y t e a c h e r$ $b l a m e d m y a n s w e r$
Commented byarkanmath7@gmail.com last updated on 19/Feb/20

$c o n f u s e d q u e s t$ $i t h i n k y o u a r e t h e t r u e s t$
Commented byjagoll last updated on 19/Feb/20

$h a h a . . . i f t h e m o s t c o r r e c t,$ $s u r e l y G o d s i r$
Commented bymr W last updated on 19/Feb/20

$b e v e r y c a r e f u l w i t h s q u a r i n g b y$ $i n e q u l i t y! i t m a y c h a n g e t h e v a l i d i t y$ $r a n g e o f t h e o r i g i n a l i n e q u a l i t y!$
Commented byjagoll last updated on 19/Feb/20

$p l e a s e s i r p o s t y o u r s t e p s o l u t i o n$
Commented byjagoll last updated on 19/Feb/20

Commented byjagoll last updated on 19/Feb/20

$i t s i r b y g r a p h i c$
Commented bymathmax by abdo last updated on 19/Feb/20

$t h e i n e q u a t i o n i s d e f i n e d f o r x ⩾ 2 a n d x^{2} - 3 x - 4 ⩾ 0$ $Δ = {(- 3)}^{2} - 4 (- 4) = 9 + 16 = 25 \Rightarrow x_{1} = \frac{3 + 5}{2} = 4 a n d x_{2} = \frac{3 - 5}{2} = - 1$ $x^{2} - 3 x - 4 ⩾ 0 \Rightarrow x \in] - \infty, - 1 [\cup] 4, + \infty [\Rightarrow D_{i n} = [2, + \infty [$ $(i n) \Rightarrow x^{2} - 3 x - 4 > x^{2} - 4 x + 4 \Rightarrow x > 8 \Rightarrow S = [8, + \infty [$
Commented byjagoll last updated on 19/Feb/20

$s o l u t i o n x ⩽ - 1 \lor x > 8$
	Answered by mr W last updated on 19/Feb/20

	$s u c h t h a t \sqrt{x^{2} - 3 x - 4} i s d e f i n e d,$ $x^{2} - 3 x - 4 = (x + 1) (x - 4) ⩾ 0$ $\Rightarrow x ⩽ - 1 o r x ⩾ 4$ $c a s e 1 : x ⩽ - 1$ $w i t h x ⩽ - 1, w e h a v e x - 2 ⩽ - 3 < 0$ $b u t \sqrt{x^{2} - 3 x - 4} ⩾ 0, s o$ $\sqrt{x^{2} - 3 x - 4} > x - 2 i s t r u e .$ $\Rightarrow x ⩽ - 1 i s s o l u t i o n .$ $c a s e 2 : x ⩾ 4$ $w i t h x ⩾ 4, x - 2 ⩾ 6 > 0$ $x^{2} - 3 x - 4 > {(x - 2)}^{2}$ $x^{2} - 3 x - 4 > x^{2} - 4 x + 4$ $x > 8 ⩾ 4$ $\Rightarrow x > 8 i s s o l u t i o n .$ $(w e c a n u s e s q u a r i n g h e r e b e c a u s e w e$ $h a v e e n s u r e d a t f i r s t t h a t x - 2 > 0)$ $s u m m a r y o f a l l s o l u t i o n s :$ $x ⩽ - 1$ $x > 8$
	Commented byjagoll last updated on 19/Feb/20

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