solve (√(2x+3))−(√(3x+8))=(√(3x+4))−(√(2x+7))


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Question Number 196056 by mr W last updated on 17/Aug/23

$s o l v e$ $\sqrt{2 x + 3} - \sqrt{3 x + 8} = \sqrt{3 x + 4} - \sqrt{2 x + 7}$
	Answered by cortano12 last updated on 17/Aug/23

	$\sqrt{u} - \sqrt{v + 4} = \sqrt{v} - \sqrt{u + 4}$ $\sqrt{u} - \sqrt{v} = \sqrt{v + 4} - \sqrt{u + 4}$ $u + v - 2 \sqrt{uv} = u + v + 8 - 2 \sqrt{(u + 4) (v + 4)}$ $\sqrt{(u + 4) (v + 4)} = 4 + \sqrt{uv}$ $uv + 4 (u + v) + 16 = 16 + uv + 8 \sqrt{uv}$ $u + v = 2 \sqrt{uv}$ $(\sqrt{u} - \sqrt{v}) = 0$ $\sqrt{u} = \sqrt{v}; [{\begin{cases} u = 2 x + 3 \\ v = 3 x + 4 \end{cases}]$ $\Leftrightarrow 3 x + 4 = 2 x + 3$ $\Leftrightarrow x = - 1$
	Commented by mr W last updated on 17/Aug/23

	$v e r y n i c e!$
	Answered by mr W last updated on 17/Aug/23

	$a n o t h e r w a y :$ $\sqrt{2 x + 7} + \sqrt{2 x + 3} = \sqrt{3 x + 8} + \sqrt{3 x + 4} . . . (i)$ $\frac{(\sqrt{2 x + 7} + \sqrt{2 x + 3}) (\sqrt{2 x + 7} - \sqrt{2 x + 3})}{\sqrt{2 x + 7} - \sqrt{2 x + 3}} = \frac{(\sqrt{3 x + 8} + \sqrt{3 x + 4}) (\sqrt{3 x + 8} - \sqrt{3 x + 4})}{\sqrt{3 x + 7} - \sqrt{3 x + 4}}$ $\frac{4}{\sqrt{2 x + 7} - \sqrt{2 x + 3}} = \frac{4}{\sqrt{3 x + 7} - \sqrt{3 x + 4}}$ $\Rightarrow \sqrt{2 x + 7} - \sqrt{2 x + 3} = \sqrt{3 x + 8} - \sqrt{3 x + 4} . . . (i i)$ $(i) + (i i) :$ $2 \sqrt{2 x + 7} = 2 \sqrt{3 x + 8}$ $2 x + 7 = 3 x + 8$ $\Rightarrow x = - 1 ✓$
	Commented by MM42 last updated on 17/Aug/23

	$a n d y o u r s o l u t i o n v e r y n i c e$
	Answered by MM42 last updated on 17/Aug/23

	$A n d o t h e r w a y$ $2 x + 3 + 3 x + 8 - 2 \sqrt{6 x^{2} + 25 x + 24} = 3 x + 4 + 2 x + 7 - 2 \sqrt{6 x^{2} + 29 x + 28}$ $\Rightarrow \sqrt{6 x^{2} + 25 x + 24} = \sqrt{6 x^{2} + 29 x + 28}$ $\Rightarrow x = - 1 ✓$
	Answered by Rasheed.Sindhi last updated on 17/Aug/23

	$\underline{\sqrt{2 x + 3} - \sqrt{3 x + 8} = \sqrt{3 x + 4} - \sqrt{2 x + 7}}$ $\sqrt{2 x + 5 - 2} + \sqrt{2 x + 5 + 2} = \sqrt{3 x + 6 - 2} + \sqrt{3 x + 6 + 2}$ $2 x + 5 = a, 3 x + 6 = b$ $\sqrt{a - 2} + \sqrt{a + 2} = \sqrt{b - 2} + \sqrt{b + 2} = c (s a y)$ $\sqrt{a - 2} + \sqrt{a + 2} = c . . . . . . . . . (i)$ ${(\sqrt{a - 2})}^{2} - {(\sqrt{a + 2})}^{2} = (a - 2) - (a + 2) = - 4 . . . (i i)$ $(i i) / (i) :$ $\sqrt{a - 2} - \sqrt{a + 2} = \frac{- 4}{c} . . . . . . . (i i i)$ $(i) + (i i i) :$ $2 \sqrt{a - 2} = c - \frac{4}{c} . . . . . (i v)$ $S i m i l a r l y,$ $2 \sqrt{b - 2} = c - \frac{4}{c} . . . . . (v)$ $(i v) & (v) : 2 \sqrt{a - 2} = 2 \sqrt{b - 2}$ $a = b \Rightarrow 2 x + 5 = 3 x + 6$ $x = - 1$
	Answered by Frix last updated on 17/Aug/23

	$Just guessing . If {there}^{'} s a ‘ ‘ {nice}^{″} solution$ $2 x + 3 = 3 x + 4 \land 3 x + 8 = 2 x + 7$ $x = - 1 \land x = - 1$
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