square root of 4y +12 - square root of y-6 = 6

Question

Its a root problem I'm having trouble understanding

 

Answer 1

√(4y + 12) - √(y - 6) = 6

 

 

The goal here is to reduce the number of radicals by applying basic algebraic concepts.

 

 

Add √(y - 6) on both sides of equation.

 

√(4y + 12) = 6 + √(y - 6)

 

 

Square both sides of equation.

 

4y + 12 = (6 + √(y - 6))²

 

4y + 12 = (6 + √(y - 6))(6 + √(y - 6))

 

4y + 12 = 36 + 12√(y - 6) + y - 6

 

4y + 12 = 30 + 12√(y - 6) + y

 

 

Subtract y and subtract 30 on both sides of equation to isolate the radical term.

 

3y - 18 = 12√(y - 6)   ----->  We only have one radical term instead of 2.

 

 

Now, we have to square both sides of equation once again.

 

(3y - 18)² = (12√(y - 6))²

 

(3y - 18)² = (12√(y - 6))(12√(y - 6))

 

9y² - 108y + 324 = 144(y - 6)     -------->  We have NO radical terms. The equation is simpler to solve now.

 

9y² - 108y + 324 = 144y - 864

 

 

Subtract 144y and add 864 on both sides of equation.

 

9y² - 252y + 1188 = 0    --------->  Quadratic equation.

 

 

Using the quadratic formula:

 

y = (252 ± √(63504 -4(10692))) / 18

y = (252 ± 144) / 18

 

y = (252 + 144) / 18                and             y = (252 - 144) / 18

 

y = 22                                     and             y = 6

 

 

If we substitute both solutions into the equation, both will give the result of 6.

 

 

Answer 2

y=6