The result is x - 2 = 81. This equation is simple to solve. We have x = 83.

In this case we still begin by squaring both sides of the equation. The result is . To finish solving this requires us to set all terms equal to zero and either factor or use the quadratic formula. We get . This factors: and the solutions are x = 2 or x = - 1. We must check each of these solutions in the original equation to see if the value of x gives a solution. x = 2 gives which is correct. x = - 1 gives which is impossible since for n > 0 is always positive. Therefore the only solution is x = 2.

x = 2 gives which is correct. x = - 1 gives which is impossible since for n > 0 is always positive.

Squaring both sides of the equation gives us . Setting terms equal to zero gives . The expression factors: . Setting each factor equal to 0 gives two possible answers: x = 3 or x = – 1. We check each answer in the original equation: If x = 3 we have which is correct. If x = – 1 we have which is impossible since the square root cannot be negative. Therefore the only answer is x = 3.

If x = 3 we have which is correct. If x = – 1 we have which is impossible since the square root cannot be negative.

We begin again by squaring both sides of the equation. The result is 3x – 2 = x + 7. We solve for x getting 2x = 9 or . We check this possible answer in the original equation. If we have . This is the same as which is correct. Therefore the solution is .

If we have . This is the same as which is correct.

We square both sides. This gives x + 2 = 2 – x. Solving for x gives 2x = 0. The only solution is x = 0. Checking this in the original equation gives which is correct. Therefore the solution is x = 0.

Checking this in the original equation gives which is correct.

Our first step is to separate the radicals. The result is . Now we square both sides. The result is . We next isolate the radical term on the right side. The result is . We square both sides again. The result is . We set the terms equal to zero. The result is . We solve this by factoring or using the quadratic formula. This does factor: (x – 5)(x + 3) = 0. This gives two possible solutions: x = 5 or x = – 3. We check these possible solutions in the original equation. If x = 5 we have . This is the same as 4 – 3 = 1. If x = – 3 we have . This is impossible since . Therefore the only solution is x = 5.

If x = 5 we have . This is the same as 4 – 3 = 1. If x = – 3 we have . This is impossible since .