I like to use a graphing calculator to solve equations involving roots. That helps avoid extraneous solutions. Here, the graph of the left side intersects the graph of the right side at x=4. There is exactly one solution, so the solution set is ...

{ 4 }

Solving this algebraically, you would isolate the radical, then "undo" it by squaring both sides of the equation. Then solve the resulting quadratic. That will have 2 solutions, only one of which will work in the original equation.

√(5x +16) -2 = 3x -8 . . . . given

√(5x +16) = 3x -6 . . . . . . add 2

5x +16 = (3x -6)^2 . . . . . square both sides

5x +16 = 9x^2 -36x +36 . . . . expand the square

9x^2 -41x +20 = 0 . . . . . . subtract (5x+16) to put into standard form

Factors of 9·20 = 180 that have a sum of 41 are 5 and 36, so we can rewrite this equation as ...

9x^2 -36x -5x +20 = 0

9x(x -4) -5(x -4) = 0 . . . . . . factor by pairs

(9x -5)(x -4) = 0

The values of x that make this equation true are the values that make the factors be zero: 5/9 and 4. Trying these in the equation, we find ...

√(5(5/9) +16) -2 = 13/3 -2 = 2 1/3 . . . . value of the left side for x = 5/9

3(5/9) -8 = -6 1/3 . . . . . . value of the right side for x=5/9; not the same

x = 5/9 is NOT A SOLUTION

√(5(4)+16) -2 = √36 -2 = 4 . . . . value of the left side for x = 4

3(4) -8 = 12 -8 = 4 . . . . . . value of the right side for x=4. These are the same, so ...

x = 4 is a solution

The set of solutions is then ...

{ 4 } . . . . . solution set

5 0

2 years ago