SOLVING EQUATIONS WITH EXTRANEOUS SOLUTIONS
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.
Problem 1 :
Solve x+1 = √(7x+15)
Solution :
x+1 = √(7x+15)
Take squares on both sides.
(x+1)2 = (√7x+15)2
Using algebraic identity (a + b)2, we get
x2 + 12 + 2(x)(1) = 7x + 15
x2 + 1 + 2x = 7x + 15
By combining the like terms, we get
x2 + 1 + 2x - 7x – 15 = 0
x2 – 5x – 14 = 0
(x – 7)(x + 2) = 0
x = 7, x = -2
|
when x = 7 7 + 1 = √(7(7) + 15) 8 = √(49 + 15) 8 = √64 8 = 8 So, 7 is a solution |
when x = -2, -2 + 1 = √(7(-2) + 15) -1 = √(-14 + 15) -1 ≠ √1 -2 is not solution |
Hence -2 is the extraneous solution.
Problem 2 :
Solve √(x + 2) + 1= √(3 – x)
Solution :
√(x + 2) + 1 = √(3 – x)
Take squares on both sides.
[√(x + 2) + 1]2 = (√(3 – x))2
(√(x + 2))2 + 2√(x + 2) (1) + 12 = 3 - x
x + 2 + 2√(x + 2) + 1 = 3 - x
2√(x + 2) + x + 3 = 3 - x
Grouping the like terms, we get
2√(x + 2) = 3 - x - x - 3
2√(x + 2) = - 2x
√(x + 2) = - x
Take squares on both sides, we get
x + 2 = x2
x2 - x - 2 = 0
(x - 2) (x + 1) = 0
x = 2 and x = -1
|
When x = -1 √(-1+2)+1= √(3–(-1) √1+1 = √(3+1) 1+1 = √4 2 = √4 |
When x = 2 √(2+2)+1= √(3–2) √4+1 = √1 Decomposing into two branches 2+1 ≠ √1 (or) -2+1 ≠ √1 |
So, 2 is the extraneous solution.
Problem 3 :
√(10 - x) = x + 2
Solution :
√(10 - x) = x + 2
Taking squares on both sides, we get
10 - x = (x + 2)2
10 - x = x2 + 2x(2) + 22
10 - x = x2 + 4x + 4
x2 + 4x + 4 + x - 10 = 0
x2 + 5x - 6 = 0
(x + 6) (x - 1) = 0
x = -6 and x = 1
To check which is solution and which is not,
|
When x = -6 √(10 - (-6)) = -6 + 2 √16 = -4 4 ≠ -4 |
When x = 1 √(10 - 1) = 1 + 2 √9 = 3 3 = 3 |
So, -6 is not a solution.
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