Solve-2x-7-x-1-2-

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 \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 \dots . (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 \dots (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