What is x if (2x+48)^1/2-x=0 ?What is x if (2x+48)^1/2-x=0 ?
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
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
giorgiana1976 | College Teacher | (Level 3) Valedictorian
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}.