Question

$\sqrt{2x + 6} - \sqrt{x + 1} = 2$
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Answer 1

${} \huge \mathfrak \green{solution - }$

$\sqrt{2x + 6} - \sqrt{x + 1} = 2 \\ \sqrt{2x + 6} = 2 + \sqrt{x + 1} \\ squaring \: both \: side - \\ 2x + 6 = 4 + (x + 1) + 2 \times 2 \sqrt{x + 1}$

$2x + 6 = 4 + x + 1 + 4 \sqrt{x + 1} \\ 2x - x + 6 - 5 = 4 \sqrt{x + 1} \\ x + 1 = 4 \sqrt{x + 1} \\ again \: squaring \: both \: side - \\ {x}^{2} + 1 + 2x = 16(x + 1) \\ {x}^{2} + 1 + 2x = 16x + 16$

${x}^{2} - 15 - 14x = 0 \\ {x}^{2} - 14x - 15 = 0 \\ {x}^{2} - (15 - 1)x - 15 = 0 \\ {x}^{2} - 15x + x - 15 = 0 \\ x(x - 15) + 1(x - 15) = 0 \\ (x + 1)(x - 15) \\ x = - 1 \\ x = 15$