# sqrt(x^2 -5) = sqrt( 2x +3) find x.

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### 3 Answers

sqrt(x^2 -5) = sqrt( 2x +3)

First we will square both sides:

[sqrt(x^2 - 5)]^2 = [ sqrt(2x+3)]^2

==> (x^2 -5) = (2x+3)

Now let us move all terms to the left side:

==> x^2 -2x -5 -3 = 0

==> x^2 - 2x - 8 = 0

Now factor theequation:

==> (x-4) (x+2) = 0

**==> x1= 4 **

**==> x2= -2 **

step 1: square each side

result of step one is: (x^2-5)=(2x+3)

step 2: you move 2x+3 to the side of x^2-5

result of step two is: x^2-2x-8

step 3: you factor out x^2-2x-8

result of step three: (x-4)(x+2)

which gives you x=4 and x=-2

sqrt(x^2-5) = sqrt(2x+3).

To find x.

Solution:

Since both sides are under suare root sign , we first get rid of the square root . So we square both sides, and then solve the resulting quadratic equation.

x^2-5 = 2x+3.

Subtract 2x+3 from both sides:

x^2-5-2x-3 = 0.

x^2-2x -8 = 0.

x^2-4x +2x-8 = 0.

x(x-4) +2(x-4) = 0.

(x-4)(x+2) = 0.

x-4 = 0. Or x+2 = 0.

x-4 = 0 gives x = 4.

x+2 = 0 gives x = -2.

Therefore the solution of the given equation is x = 4 or x = -2.