Solve the equation
Solution:Square both sides of the original equation to obtain the equation x² = -5x - 6.
Note:This equation is not equivalent to the original equation.
However, the solution set of this equation contains the solution set of the first equation.
Add 5x + 6 to both sides ot the second equation to obtain x² + 5x + 6 = 0.
Factorization yields (x + 2)(x + 3) = 0 and then The Zero Factor Property yields the two equations
x + 2 = 0 OR x + 3 = 0. The solution sets to these two equations are {-2} OR {-3}.
The solution set of the second equation x² = -5x - 6 is therefore .
Every solution of the original equation is contained in the set {-2, -3}, but some of the elements of the set {-2, -3} may not be solutions of the original equation. Therefore we must test each of these possible solutions.
Test -2:  
Therefore -2 is not a solution of the original equation.
Test -3:  
Therefore -3 is not a solution of the original equation.
Neither of the two possible solutions satisfy the original equation.
Therefore the solution set for the original equation is the empty set

Observe that

Solve the equation
Solution:Square both sides of the original equation to obtain the equation 2x + 9 = x² + 6x +9.
Note:This equation is not equivalent to the original equation.
However, the solution set of this equation contains the solution set of the first equation.
Add -2x - 9 to both sides ot the second equation to obtain x² + 4x = 0.
Factorization yields x (x + 4) = 0 and then The Zero Factor Property yields the two equations
x = 0 OR x + 4 = 0. The solution sets to these two equations are {0} OR {-4}.
The solution set of the second equation 2x + 9 = x² + 6x +9. is therefore .
Every solution of the original equation is contained in the set {0, -4}, but some of the elements of the set {0, -4} may not be solutions of the original equation. Therefore we must test each of these possible solutions.
Test 0:  
Therefore 0 is a solution of the original equation.
Test -4:  
Therefore -4 is not a solution of the original equation.
Therefore the solution set for the original equation is the set {0}.

Observe that

Solve the equation
Solution:Square both sides of the original equation to obtain the equation x² = 5x - 6.
Note:This equation is not equivalent to the original equation.
However, the solution set of this equation contains the solution set of the first equation.
Add -5x + 6 to both sides ot the second equation to obtain x² - 5x + 6 = 0.
Factorization yields (x - 2)(x - 3) = 0 and then The Zero Factor Property yields the two equations
x - 2 = 0 OR x - 3 = 0. The solution sets to these two equations are {2} OR {3} respectively.
The solution set of the second equation x² - 5x + 6 = 0. is therefore .
Every solution of the original equation is contained in the set {2, 3}, but some of the elements of the set {2, 3} may not be solutions of the original equation. Therefore we must test each of these possible solutions.
Test 2:  
Therefore 2 is a solution of the original equation.
Test 3:  
Therefore 3 is a solution of the original equation.
We conclude the solution set for the original equation is the set {2, 3}.

Observe that