Solve (√(2x+7)) −(√(x−1)) =2


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Question Number 217940 by ArshadS last updated on 23/Mar/25

$S o l v e$ $\sqrt{2 x + 7} - \sqrt{x - 1} = 2$
	Answered by Frix last updated on 23/Mar/25
	$No real solution because min (lhs) >2 a, b ∈R∧b≠0 (√(2x+7))=a+2+bi ⇒ x=((a^2 −b^2 +4a−3)/2)+(a+2)bi (√(x−1))=a+bi ⇒ x=a^2 −b^2 +1+2abi ⇒ { ((a^2 −b^2 +4a−3=2(a^2 −b^2 +1))),(((a+2)b=2ab)) :} { ((a^2 −b^2 −4a+5=0)),((a=2 [b≠0])) :} { ((b=±1)),((a=2)) :} ⇒ x=4±4i$
	$No real solution because \min (lhs) > 2$ $a, b \in R \land b \neq 0$ $\sqrt{2 x + 7} = a + 2 + b i \Rightarrow x = \frac{a^{2} - b^{2} + 4 a - 3}{2} + (a + 2) b i$ $\sqrt{x - 1} = a + b i \Rightarrow x = a^{2} - b^{2} + 1 + 2 a b i$ $\Rightarrow$ ${\begin{cases} a^{2} - b^{2} + 4 a - 3 = 2 (a^{2} - b^{2} + 1) \\ (a + 2) b = 2 a b \end{cases}$ ${\begin{cases} a^{2} - b^{2} - 4 a + 5 = 0 \\ a = 2 [b \neq 0] \end{cases}$ ${\begin{cases} b = \pm 1 \\ a = 2 \end{cases}$ $\Rightarrow x = 4 \pm 4 i$
	Answered by mehdee7396 last updated on 23/Mar/25

	$\sqrt{2 x + 7} = 2 + \sqrt{x - 1}$ $2 x + 7 = 4 + 4 \sqrt{x - 1} + x - 1$ $4 \sqrt{x - 1} = - x - 4$ $16 x - 16 = x^{2} + 8 x + 16$ $x^{2} - 8 x + 32 = 0$ $x = 4 \pm \sqrt{- 16} = 4 \pm 4 i$
	Answered by Rasheed.Sindhi last updated on 24/Mar/25

	$\sqrt{2 x + 7} - \sqrt{x - 1} = 2 . . . . (i)$ $(\sqrt{2 x + 7} - \sqrt{x - 1}) (\sqrt{2 x + 7} + \sqrt{x - 1}) = 2 (\sqrt{2 x + 7} + \sqrt{x - 1})$ $2 (\sqrt{2 x + 7} - \sqrt{x - 1}) = 2 x + 7 - x + 1 = x + 8$ $\sqrt{2 x + 7} + \sqrt{x - 1} = \frac{x}{2} + 4 . . . (i i)$ $(i) + (i i) : 2 \sqrt{2 x + 7} = \frac{x}{2} + 6$ $4 \sqrt{2 x + 7} = x + 12$ $16 (2 x + 7) = x^{2} + 24 x + 144$ $x^{2} - 8 x + 32 = 0$ $x = \frac{8 \pm \sqrt{64 - 128}}{2} = 4 \pm 4 i$
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