Question 62331: sqrt(3x-5) - sqrt(x+7)=2 solve for x...may be more than one answer
Found 3 solutions by jai_kos, MathTherapy, ikleyn:
Answer by jai_kos(139)   (Show Source):
You can put this solution on YOUR website! Given sqrt(3x-5) - sqrt(x+7)= 2
==> sqrt(3x -5) = 2 + sqrt( x+7)
Square on both sides, we get
( 3x -5) = [ 2 + sqrt(x + 7)] ^2
3x - 5 = 4 + ( x + 7) + 4sqrt( x + 7)
3x - 5 - 4 -x - 7 = 4sqrt ( x + 7)
2x - 16 = 4 sqrt(x +7)
Divide throughout by 2, we get
x - 8 = 2sqrt (x + 7)
Square on both sides, we get
( x -8) ^ 2 = 4 ( x + 7)
x ^2 - 16x + 64 = 4x + 28
x ^2 - 16x -4x + 64 - 28 = 0
x^2 - 20x + + 36 = 0
x^2 - 18x - 2x + 36 = 0
x (x - 18) - 2 (x - 18) = 0
( x - 2) ( x- 18) = 0
==> x - 2 = 0 or x - 18 = 0
==> x = 2 or x = 18
Answer by MathTherapy(10801)  (Show Source):
You can put this solution on YOUR website!
sqrt(3x-5) - sqrt(x+7)=2 solve for x...may be more than one answer
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According to whomever responded, x has 2 values, 18 and 2. However,
only 18 satisfies the given equation. The other value, 2, is EXTRANEOUS!
Answer by ikleyn(53742)  (Show Source):
You can put this solution on YOUR website! .
sqrt(3x-5) - sqrt(x+7)=2 solve for x...may be more than one answer
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The answer in the post by @jai_kos, giving two solutions x= 2 and x= 18, is incorrect.
Of two possible values, x= 2 and x= 18, only x= 18 is the actual solution.
Other value, x= 2, is EXTRANEOUS solution, which does not satisfies the original equation
This extraneous solution should be rejected at the checking process, which @jai_kos neglected to make.
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