Solve the equation (x-4)^1/2=1/(x-4).Solve the equation (x-4)^1/2=1/(x-4).  

Asked on by undoitu

3 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

We’ve answered 319,865 questions. We can answer yours, too.

Ask a question