Solve the equation (x-4)^1/2=1/(x-4).Solve the equation (x-4)^1/2=1/(x-4).
3 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
The equation to be solved is (x-4)^1/2=1/(x-4)
(x-4)^1/2=1/(x-4)
square both the sides
=> x - 4 = 1 / (x - 4)^2
=> (x - 4)^3 = 1
=> 1 - (x - 4)^3 = 0
=> (1 - (x - 4))(1 + x - 4 + (x - 4)^2) = 0
=> (1 - x + 4)(1 + x - 4 + x^2 + 16 - 8x) = 0
=> (-x + 5)(x^2 - 7x + 13) = 0
-x + 5 = 0
=> x = 5
x^2 - 7x + 13 = 0
=> x1 = 7/2 + sqrt(49 - 52) /2
=> x1 = 7/2 + i*(sqrt 3)/2
x2 = 7/2 - i*(sqrt 3/2)
The equation has the solutions x = (5, 7/2 + (sqrt 3)/2, 7/2 - (sqrt 3/2))
giorgiana1976 | College Teacher | (Level 3) Valedictorian
We'll impose the constraints of existence of the square root:
x - 4> =0
x>=4
Now, we'll solve the equation by raising to square both sides, to eliminate the square root:
(x - 4) = 1/(x - 4)^2
Now, we'll subtract 1/(x - 4)^2 both sides:
(x - 4) - 1/(x - 4)^2 = 0
We'll multiply by (x-4)^2 the equation:
(x - 4)^3 - 1>=0
We'll solve the difference of cubes using the formula:
a^3 - b^3 = (a-b)(a^2 + ab + b^2)
(x - 4)^3 - 1 = (x - 4 -1)[(x-4)^2 + x - 4 + 1]
We'll combine like terms inside brackets:
(x - 5)[(x-4)^2 + x - 4 + 1] = 0
We'll cancel each factor:
x - 5 = 0
We'll add 5 both sides:
x = 5
(x-4)^2 + x - 4 + 1 = 0
We'll expand the square:
x^2 - 8x + 16 + x - 3 = 0
We'll combine like terms:
x^2 - 7x + 13 = 0
We'll apply quadratic formula:
x1 = [7 + sqrt(49 - 52)]/2
x1 = (7 + i*sqrt3)/2
x2 = (7 - i*sqrt3)/2
The roots of the given equation are complex numbers. Since there is not any constraint concerning the nature of roots, we'll accept them.
pinkgirly | Student, Grade 11 | (Level 1) eNoter
(x-4)^(1/2) = 1/ (x-4)
√(x-4) = 1/(x-4)
x-4 =1
x=5