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

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve for x given that (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 - 8)(x + 6) = 0

From x - 8 = 0, we get x = 8

and from x + 6 = 0, we get x = -6

Therefore x can take the values -6 and 8.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll re-write the equation, adding x both sides:

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 range 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

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

x1 = (2+14)/2

x1 = 8

x2 = (2-14)/2

x2 = -6

We notice that both values belong to the range of admissible values, so we'll validate them. The solutions of the equation are {-6 ; 8}.