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Find a negative value of x that satisfies the equation [(x + 1)^{2} - (2x + 1)]^{1/2} + 2|x| - 6 = 0

A) -5

B) -4

C) -3

D) -2

E) -1

**Comments**

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[(x + 1)^{2 }- (2x + 1)]^{1/2} + 2|x| - 6 = 0

From expansion,

(x + 1)^{2} = (x + 1)(x+1)

= x^{2} + 2x + 1

Substitute this into the initial equation

[x^{2} + 2x + 1 -(2x + 1)]^{1/2} + 2|x| - 6 = 0

[x^{2} + 2x + 1 -2x - 1]^{1/2} + 2|x| - 6 = 0

[x^{2}]^{1/2} + 2|x| - 6 = 0

And, x^{1/2} = √x, Similarly (x^{2})^{1/2} = √x^{2} = x

x + 2|x| = 6

3x = 6

x = 6/3

x = 2

But since we want a negative value

x = -2

**Ans D**

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