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

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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First, 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:

(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 put each factor as zero:

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 imposed with regard to the nature of roots, we'll accept them.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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

We square both sides of the given equation:

(x-4) = 1/(x-4)^2.

We multiply both sides by (x-4)^2.

(x-4)^3 = 1.

We take cube roots of both sides to obtain real roots.

x-4 = 1.

We add 4 to both sides:

x = 1+4 = 5.

Therefore the solution is x = 5.

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