HOW TO FIND EXTRANEOUS SOLUTION IN RADICAL EQUATIONS
Extraneous solutions are values that we get when solving equations that aren't really solutions to the equation.
Difference between solution and extraneous solution :
The value or values which satisfies the given equation is known as solution.
The value or values which are not satisfying the given equation is known as extraneous solution.
Let a = b
Take squares on both sides.
Then,
a2 = b2
But vice versa,
If a2 = b2, then a and b need not to be true.
To solve radical equations, we follow the steps given below.
- To remove the radical sign, based on the index of the radical, raise the powers on both sides. For example,
- If we have square root on both sides, we have to take squares on both sides.
- If we have cube root on both sides, we have to take cubes on both sides.
- Sometimes we may have to use algebraic identities to find the expansion.
- Then we solve the equation.
Find the extraneous solution for the radical equations given below.
Problem 1 :
√(x + 4) = x - 2
Solution :
√(x+4) = x-2
Take power 2 on both sides.
[√(x + 4)]2 = (x-2)2
x + 4 = x2 - 2x(2) + 22
x + 4 = x2 - 4x + 4
x2 - 4x + 4 - x - 4 = 0
x2 - 5x = 0
x(x - 5) = 0
x = 0 and x = 5
|
When x = 0 √(0+4) = 0-2 √4 ≠ -2 2 ≠ -2 0 is the solution. |
When x = 5 √(5+4) = 5-2 √9 = 3 3 = 3 5 is not the solution. |
So, 5 is extraneous solution.
Problem 2 :
√(10x + 9) = x + 3
Solution :
√(10x+9) = x+3
Take power 2 on both sides.
[√(10x+9)]2 = (x+3)2
10x + 9 = x2 + 32 + 2(x)(3)
10x + 9 = x2 + 9 + 6x
x2 + 6x – 10x + 9 – 9 = 0
x2 - 4x = 0
x(x – 4) = 0
x = 0, x = 4
√(10x + 9) = x + 3
|
When x = 0 √(10(0) + 9) = 0 + 3 √9 = 3 3 = 3 0 is the solution. |
When x = 4 √(10(4) + 9) = 4 + 3 √40 + 9 = 7 √49 = 7 7 = 7 x = 4 is a solution |
So, there is no extraneous solution.
Problem 3 :
√(2x + 5) = √(x + 7)
Solution :
√(2x + 5) = √(x + 7)
Take power 2 on both sides.
(√2x + 5)2 = (√x + 7)2
2x + 5 = x + 7
2x + 5 – x - 7 = 0
x – 2 = 0
x = 2
√(2x + 5) = √(x + 7)
When x = 2,
√(2(2) + 5) = √(2 + 7)
√(4 + 5) = √9
√9 = √9
3 = 3
So, there is no extraneous solution.
Problem 4 :
√(x + 6) – 2 = √(x – 2)
Solution :
√(x + 6) – 2 = √(x – 2)
Take power 2 on both sides.
[√(x + 6) – (2)]2 = (√(x – 2))2
(√x + 6)2 + (2)2 – 2(√(x + 6))(2) = = (√(x – 2))2
x + 6 + 4 - 4√(x + 6) = (x – 2)
x + 10 - 4√(x + 6) = x – 2
x + 10 - 4√(x + 6) - x + 2 = 0
12 - 4√(x + 6) = 0
12 = 4√(x + 6)
12/4 = √(x + 6)
3 = √(x + 6)
Take power 2 on both sides.
32 = (√(x + 6))2
9 = x + 6
9 – 6 = x
3 = x
√(x + 6) – 2 = √(x – 2)
When x = 3,
√(3 + 6) – 2 = √(3 – 2)
√9 – 2 = √1
3 - 2 =√1
1 = 1
x = 3 is satisfies the given equation. So, no extraneous solution.
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