Rebecca L.

asked • 11/23/14

radical equation x+1+5=x

How would I solve this radical equation with extraneous solutions?

Philip P.

tutor
As written, the equation is unsolvable as Christopher points out.    You call it a radical equation, but there is no radical.  If there is a radical, please correct the equation and show where it goes.
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11/23/14

Philip P.

tutor
For instance, if the equation were:
 
√(x+1) + 5 = x
√(x+1) = x - 5
 
Square both sides:
 
x + 1 = x2 - 10x +25
0 = x2 - 11x + 24
 
Factors to:
 
0 = (x-3)(x-8)
x = 3, 8
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11/23/14

1 Expert Answer

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Christopher R. answered • 11/23/14

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Rebecca, there are some mathematical problems that have no solution. The equation x+1+5=x has no solution. The key to proving this is to simplify the equation.
 
x+1+5=x+6=x
           -x     -x   
 
6≠0  Hence, the answer to the solution to this equation is there is no value for x to make the equation true. In a nutshell, there's no solution.
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Christopher R.

You could pick any value for x and it still wouldn't make the equation true.
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11/23/14

Christopher R.

Here's some food for thought. consider this equation and solve for x.
 
 ax+6=bx
-ax     -ax
       6=(b-a)x
 
x=6/(b-a)   This implies a≠b otherwise, the solution becomes undefined because you have a number divided by zero.
 
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11/23/14

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