# sqrt(x^2 + 2) = x+3 find x value.

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sqrt(x^2 + 2) = x+ 3

First let us square both sides:

==> [sqrt(x^2 + 2)]^2 = (x+3)^2

==> (x^2 + 2) = x^2 + 6x + 9

Now let us group similar terms:

==> x^2 + 2 - x^2 - 6x -9 0

==> -6x - 7 = 0

Now add 7 to both sides:

==> -6x = 7

Now divide by -6:

**==> x= -7/6**

We'll start by imposing constraints of existence of the square root.

x^2 + 2 > 0

Since te value of x^2 is always positive, no matter the value of x is, the expression x^2 + 2 > 0.

Now, we'll square raise both sides to get rid of the square root.:

[sqrt(x^2 + 2)]^2 = (x+3)^2

We'll expand the square from the right side:

x^2 + 2 = x^2 + 6x + 9

We'll subtract x^2 + 6x + 9 and we'll eliminate like terms:

-6x - 9 + 2 = 0

-6x - 7 = 0

We'll add 7 both sides:

-6x = 7

We'll divide by -6:

x = -7/6

**x = -1.1(6)**