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Question Number 22313 by math solver last updated on 15/Oct/17

Commented by Joel577 last updated on 16/Oct/17
$(√(5x + 7)) − (√(3x + 1)) = (√(x + 3)) x ≥ −(7/5) and x ≥ −(1/3) and x ≥ −3 ⇒ x ≥ −(1/3) ((√(5x + 7)) − (√(3x + 1)))^2 = x + 3 5x + 7 − 2(√((5x + 7)(3x +1))) + 3x +1 = x + 3 7x + 5 − 2(√((5x + 7)(3x +1))) = 0 7x + 5 = 2(√((5x + 7)(3x + 1))) 49x^2 + 70x + 25 = 60x^2 + 104x + 28 11x^2 + 34x + 3 = 0 x_1 = −3 x_2 = −(1/(11)) Hence, the solution that satisfy the equation above is {−(1/(11))}$
$\sqrt{5 x + 7} - \sqrt{3 x + 1} = \sqrt{x + 3}$ $x ⩾ - \frac{7}{5} and x ⩾ - \frac{1}{3} and x ⩾ - 3$ $\Rightarrow x ⩾ - \frac{1}{3}$ ${(\sqrt{5 x + 7} - \sqrt{3 x + 1})}^{2} = x + 3$ $5 x + 7 - 2 \sqrt{(5 x + 7) (3 x + 1)} + 3 x + 1 = x + 3$ $7 x + 5 - 2 \sqrt{(5 x + 7) (3 x + 1)} = 0$ $7 x + 5 = 2 \sqrt{(5 x + 7) (3 x + 1)}$ $49 x^{2} + 70 x + 25 = 60 x^{2} + 104 x + 28$ $11 x^{2} + 34 x + 3 = 0$ $x_{1} = - 3$ $x_{2} = - \frac{1}{11}$ $Hence, the solution that satisfy the equation$ $above is {- \frac{1}{11}}$
Commented by math solver last updated on 16/Oct/17

$t h a n k s :)!$
	Answered by squidward last updated on 16/Oct/17

	$i t i s a q u a d r a t i c e q u a t i o n, h e n c e$ $t h e r e a r e 2 s o l u t i o n s$
	Commented by squidward last updated on 16/Oct/17

	$y o u s q u a r e b o t h s i d e s, t h e n y o u$ $r e a r r a n g e a n d s q u a r e b o t h s i d e s a g a i n$ $i t w i l l b e a q u a d r a t i c .$
	Commented by math solver last updated on 16/Oct/17

	$h o w d i d y o u s e e d i r e c t l y t h e q u a d r a t i c$ $e q u a t i o n ?$
	Commented by ajfour last updated on 16/Oct/17

	$i f y o u s q u a r e {y o u}^{'} l l h a v e t o s e e$ $i f a n o b t a i n e d r o o t i s a c c e p t a b l e$ $o r n o t .$
	Commented by math solver last updated on 16/Oct/17

	$e x a c t l y! y o u c a n t d i r e c t l y a r g u e f o r 2 s o l u$ $s o l u t i o n s . . .$
	Commented by squidward last updated on 17/Oct/17

	$y e s i c a n . i t w a s n t s p e c i f i e d i f t h e$ $s o l u t i o n s h a v e t o b e r e a l . h e n c e t h e r e$ $a r e e x a c t l y 2 s o l u t i o n s e i t h e r i m a g i n a r y$ $o r n o t$
	Commented by math solver last updated on 17/Oct/17

	$y o u m e a n t h e a n s w e r o f t h e a b o v e p r o$ $b l e m s h o u l d b e 2 t h e n ?$
	Commented by squidward last updated on 17/Oct/17

	$y e s, f i r s t r o o t i s - \frac{1}{11} a n d s e c o n d - 3$ $(c o n s i d e r i n g i m a g i n a r y s p a c e) .$ $t r y s u b s t i t u t i n g t h e m y o u r s e l f, y o u$ $w i l l s e e$
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