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Question Number 153464 by roxceefocus last updated on 07/Sep/21

Commented by roxceefocus last updated on 07/Sep/21

$p l s i n e e d a n s w e r s o n t h i s q u e s t i o n$
	Answered by Rasheed.Sindhi last updated on 07/Sep/21

	$\sqrt{3 x + 1} - \sqrt{2 x - 1} = 1$ ${(\sqrt{3 x + 1} - \sqrt{2 x - 1} = 1)}^{2}$ $3 x + 1 + 2 x - 1 - 2 \sqrt{(3 x + 1) (2 x - 1} = 1$ ${(5 x - 1 = 2 \sqrt{(3 x + 1) (2 x - 1)})}^{2}$ ${(5 x - 1)}^{2} = 4 (3 x + 1) (2 x - 1)$ $25 x^{2} - 10 x + 1 = 4 (6 x^{2} - 3 x + 2 x - 1)$ $25 x^{2} - 10 x + 1 = 24 x^{2} - 4 x - 4$ $x^{2} - 6 x + 5 = 0$ $(x - 1) (x - 5) = 0$ $x = 1 ∣ x = 5$ $O n v e r i f i c a t i o n w e c a n s e e t h a t$ $b o t h r o o t s s a t i s f y t h e o r i g i n a l e q n$
	Answered by puissant last updated on 07/Sep/21
	$x≥−(1/3) and x≥(1/2) ⇒ x≥(1/2). we have (√(3x+1))=1+(√(2x−1)) ⇒ 3x+1=1+2(√(2x−1))+2x−1 ⇒ x+1=2(√(2x−1)) ⇒ x^2 +2x+1=8x−4 ⇒ x^2 −6x+5=0 x_1 =1 and x_2 =5 S_R ={1,5}.$
	$x ⩾ - \frac{1}{3} a n d x ⩾ \frac{1}{2} \Rightarrow x ⩾ \frac{1}{2} .$ $w e h a v e \sqrt{3 x + 1} = 1 + \sqrt{2 x - 1}$ $\Rightarrow 3 x + 1 = 1 + 2 \sqrt{2 x - 1} + 2 x - 1$ $\Rightarrow x + 1 = 2 \sqrt{2 x - 1}$ $\Rightarrow x^{2} + 2 x + 1 = 8 x - 4$ $\Rightarrow x^{2} - 6 x + 5 = 0$ $x_{1} = 1 a n d x_{2} = 5$ $S_{R} = {1, 5} .$
	Answered by Rasheed.Sindhi last updated on 08/Sep/21

	$\equiv A n A lternative \equiv$ $\sqrt{3 x + 1} - \sqrt{2 x - 1} = 1$ $\Rightarrow p - q = 1$ $p^{2} - q^{2} = (3 x + 1) - (2 x - 1) = x + 2$ $(p - q) (p + q) = x + 2$ $p + q = x + 2$ $p - q = 1$ $2 p = x + 3 \Rightarrow p = \frac{x + 3}{2}$ $2 q = x + 1 \Rightarrow q = \frac{x + 1}{2}$ $p = \sqrt{3 x + 1} = \frac{x + 3}{2}$ $4 (3 x + 1) = x^{2} + 6 x + 9$ $x^{2} - 6 x + 5 = 0$ $(x - 1) (x - 5) = 0$ $x = 1, 5$ $O n v e r i f i c a t i o n i t c a n b e v e r i f i e d$ $t h a t b o t h r o o t s a r e t r u e .$ $(S a m e r e s u l t c a n b e o b t a i n e d$ $b y$ $q = \sqrt{2 x - 1} = \frac{x + 1}{2}$ $4 (2 x - 1) = x^{2} + 2 x + 1$ $x^{2} - 6 x + 5 = 0)$
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