# What is x if (2x+48)^1/2-x=0 ?What is x if (2x+48)^1/2-x=0 ?

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The equation to be solved is (2x+48)^1/2 - x = 0

(2x+48)^1/2 - x = 0

=> (2x+48)^1/2 = x

square both the sides

2x + 48 = x^2

=> x^2 - 2x - 48 = 0

=> x^2 - 8x + 6x - 48 = 0

=> x(x - 8) + 6(x - 8) = 0

=> (x + 6)(x - 8) = 0

**The solution of the equation are x = -6 and x = 8**

We'll shift x to the right, isolating the square root to the left:

sqrt(2x + 48) = x

Now, we'll have to impose the constraint of existence of the square root:

2x + 48 >= 0

We'll subtract 48:

2x >= -48

We'll divide by 2:

x >= -24

The interval of admissible values for x is: [-24 ; +infinite)

Now, we'll solve the equation by raising to square both sides:

x^2 = [sqrt(2x + 48)]^2

x^2 = 2x + 48

We'll subtract 2x + 48 both sides:

x^2 - 2x - 48 = 0

We'll apply the quadratic formula:

x1 = [2 + sqrt(4 + 192)]/2

x1 = (2+14)/2

x1 = 8

x2 = (2-14)/2

x2 = -6

**Since both values are in the range of admissible values, we'll accept them as solutions of the equation: {-6 ; 8}.**