Solve for real x the equation x=6[(x-2)^1/2-1].

3 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve the equation x = 6[(x-2)^1/2 - 1]

x = 6[(x-2)^1/2 - 1]

=> x = 6*( x - 2)^1/2 - 6

=> x + 6 = 6*( x - 2)^1/2

take the square of both the sides

=> x^2 + 36 + 12x = 36 ( x - 2)

=> x^2 + 36 + 12x - 36x + 72 = 0

=> x^2 - 24x + 108 = 0

=> x^2 - 18x - 6x + 108 = 0

=> x( x - 18) - 6( x - 18) = 0

=> (x - 6)(x - 18) = 0

Therefore x can take the values 6 and 18.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve the equation  x=6[(x-2)^1/2-1] for real roots.

x = 6(x-2)^(1/2) -6.

We add 6 to both sides:

x+6 = (x-2)^(1/2).

We square both sides:

(x+6)^2 = 6(x-2).

x^2+12x+36 = 6x-12

x^2+12x+36-6x+12 = 0.

x^2+6x+48= 0.

The discriminant = (coefficient of x)^2 - 4*coefficient of x^2* constant term = 6^2- 4*48 = 36 -192 = -156 which is negative.

If the discriminant is negative, then there are no real roots for the equation. So there are no real roots for x=6[(x-2)^1/2-1.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

For the beggining, we'll impose constraint of existence of the square root:

x - 2 >= 0

x >= 2

The real solutions of the equation have to belong to the range [2 ; +infinite).

Now, we'll solve the equation. We'll start by removing the brackets:

x = 6sqrt(x-2) - 6

We'll add 6 both sides, in order to isolate the square root to the right side.

x + 6 = 6sqrt(x-2)

Now, we'll square raise to eliminate the square root;

x^2 + 12x + 36 = 36x - 72

We'll subtract 36x - 72:

x^2 + 12x + 36 - 36x + 72 = 0

We'll combine like terms:

x^2 - 24x + 108 = 0

We'll apply quadratic formula:

x1 =[24 + sqrt(576 - 432)]/2

x1 = (24 + 12)/2

x1 = 18

x2 = (24 - 12)/2

x2 = 6

Since both values belong to the interval of admissible values, we'll accept them. The solutions of the equation are: {6 ; 18}.

We’ve answered 318,925 questions. We can answer yours, too.

Ask a question