Question

Solve $\sqrt{x +3} \leq x+1$

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Answer 1

Answer:

$\sqrt{x + 3} = x + 1$

by squaring on both sides,

${ (\sqrt{x + 3} )}^{2} = (x + 1) ^{2} \\ x + 3 = {x}^{2} + {1}^{2} + 2(x)(1)$

$[{a}^{2} + {b}^{2} + 2ab = (a + b) ^{2}]$

$x + 3 = {x}^{2} + 1 + 2x \\ {x}^{2} + 2x + 1 = x + 3 \\ {x}^{2} + 2x - x + 1 - 3 = 0 \\ {x}^{2} + 2x - x - 2 = 0$

Here we didn't subtract 2x-x because if we do factorization again we will get the same equation.

$x(x + 2) - 1(x + 2) = 0 \\ (x - 1)(x + 2) = 0 \\ here \: (x - 1) = 0 \: (or) \: (x + 2) = 0 \\ x - 1 = 0 \: (or) \: x + 2 = 0 \\ x = 1 \: (or) \: x = ( - 2)$

The value of x is 1 or (-2).

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Answer 2

Answer:

$\huge\underbrace\mathfrak\pink{Answer}$