Dirty Extraneous
Find the sum of the solutions of x of the equation:
x + x − 2 = 4
The answer is 3.
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2 solutions
as-salam mohammad al ali.....
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Alaikom asalam brother. How may i help you :) ?
What is extraneous root
What is extraneous root?
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It is a number obtained in the process of solving an equation that does not satisfy the equation.
Example of it is:
2 y + 7 + 4 = y
First, get the radical expression alone on one side of the equation: 2 y + 7 = y − 4
Squaring both sides to get rid of the radical on LHS: 2 y + 7 = ( y − 4 ) 2
Simplify the RHS: 2 y + 7 = y 2 − 8 y + 1 6
Compute y (put everything on one side): y 2 − 1 0 y + 9 = 0
Factorising it we obtain: ( y − 9 ) ( y − 1 ) = 0
Hence the solutions for y are , y = 9 , 1
But when plugging y=1 in the equation it wont work: 2 ( 1 ) + 7 + 4 = 1 is false!
Hence, y=1 is an extraneous root.
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@Amritanshu Kumar – Yes but the question has +sign with the root...and not plus-minus...
Thank you ...I tried 6 & thus failed to solve it...
thanks Mohammad Al Ali for explaining extraneous root
Expressing the equation again as ( underroot X-2)whole square =( 4-X)whole square solving this equation we get xsquare-9x+18 =0 so values of X are 3 and 6. when substituted in given equation 6 gives error results ,therefore X =3 Ans. K.K.GARG,India
why is 6 extraneous and 3 is not???
for x=6 ::
x+
x
−
2
= 4
6 +
6
−
2
= 4
6 +
(
±
2
)
= 4
so this is also true in case we take -ve value(-2) and fails in case we take +2
for x=3 ::
x+
x
−
2
= 4
3 +
3
−
2
= 4
3 + (±1) = 4
so this one also fails in case we take -ve value(-1)
Can anyone explain??
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generally we take the positive value when we have a root symbol. but if it is 9^1/2 we take it as plus or minus 3.on the other hand if it is root 9 we take the positive value i.e. just 3. there some conventions used like this.
We are given: x + \sqrt{x-2} = 4 \tag{1}
Isolating the square root on one side to get rid of it, ( x − 2 ) 2 = ( 4 − x ) 2
Hence, x − 2 = x 2 − 8 x + 1 6
Putting everything on one side we obtain, x 2 − 9 x + 1 8 = 0
Factorizing this equation we get, ( x − 3 ) ( x − 6 ) = 0 .
So, x = 3 and x = 6 . But, x = 6 is an extraneous root, and hence invalid solution (test for yourself).
Thus the only solution to ( 1 ) is x = 3 .