Squaring each side of an equation sometimes produces extraneous solutions. An extraneous solution is a solution derived from an equation that is not a solution of the original equation.
Therefore, you must check all solutions in the original equation when you solve radical equations.
Example 1 :
Solve √(x-2) = (x- 4)
Solution :
[√(x-2)]2 = (x- 4)2
(x - 2) = (x2 - 2 x (4) + 42)
x - 2 = x2 - 8 x + 16
x2 - 8 x + 16 - x + 2 = x - 2 - x + 2
x2 - 9 x + 18 = 0
(x - 6) (x - 3) = 0
x - 6 = 0 Add 6 on both sides x = 6 |
x - 3 = 0 Add 3 on both sides x = 3 |
let us apply the above values one by one to check which is an extraneous solution.
x = 6 √(6-2) = (6-4) √4 = 2 2 = 2 |
x = 3 √(3-2) = (3 - 4) √1 = -1 1 ≠ -1 |
Since 3 does not satisfy the original equation, 6 is the only solution.
Hence 3 is the extraneous solution and 6 is the solution.
Example 2 :
Solve x = √(6 - x)
Solution :
x = √(6 - x)
Take squares on both sides,
x2 = [√(6 - x)]2
x2 = 6 - x
Add x and subtract 6 on both sides, we get
x2 - 6 + x = 6 - x + x - 6
x2 + x - 6 = 0
By factoring the above quadratic equation, we get
(x - 2)(x + 3) = 0
x - 2 = 0 Add 2 on both sides, we get x - 2 + 2 = 0 + 2 x = 2 |
x + 3 = 0 Subtract 3 on both sides, we get x + 3 - 3 = 0 - 3 x = -3 |
Now let us apply each values one by one in the original equation,
x = √(6 - x) x = 2 2 = √(6 - 2) 2 = √4 2 = 2 |
x = √(6 - x) x = -3 -3 = √(-3 - 2) -3 ≠ √-5 |
Since -3 does not satisfy the original equation, 2 is the only solution.
Hence 2 is the extraneous solution and 6 is the solution.
Example 3 :
Solve x = √(x + 20)
Solution :
x = √x + 20
Take squares on both sides,
x2 = [√(x + 20)]2
x2 = x + 20
x2 - x - 20 = x + 20 - x - 20
x2 - x - 20 = 0
By factoring the above quadratic equation, we get
(x - 5)(x + 4) = 0
x - 5 = 0 Add 5 on both sides x - 5 + 5 = 0 + 5 x = 5 |
x + 4 = 0 Subtract 4 on both sides x + 4 - 4 = 0 - 4 x = -4 |
Now let us apply each values one by one in the original equation,
x = √(x + 20) x = 5 5 = √(5 + 20) 5 = √25 5 = 5 |
x = √(x + 20) x = -4 -4 = √(-4 + 20) -4 = √16 -4 ≠ 4 |
Since -4 does not satisfy the original equation, 5 is the only solution.
Hence -4 is the extraneous solution and 5 is the solution.
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