REALLY NEED HELP Solving for X with radicals/square roots

  • Thread starter Euphoriet
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  • #1
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Ok so I know that when you have radicals such as the following you only have one answer right?...


x + 1 - 2(square root of [x+4]) = 0

ok so
x = 5 right?.... i dunno why but it just does... I cant solve it

If i solve it

It works looks like this:
(squaring everything I get this in the end)

x^2-2x-15=0

ok so the answers for x = 5 ORRRRRRRRRRRR x = -3????

why is this so... I must be doing this wrong.. but I dunno how its done=/ plz help.

if I graph them... the first one just is a l ine that just "ends" =/ and the other one is a parabula of course.
 

Answers and Replies

  • #2
Tide
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Substitute both values back into the original equation and see what you get!
 
  • #3
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I know I can do that... and I know I can prove that one doesnt work BUT is there any other way of doing it.
 
  • #4
Tide
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The way you solved it is fine but you have to realize that when you squared both sides of the equation you introduced extraneous roots. In the end, the numbers that you find must satisfy the original equation!
 
  • #5
uart
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If the original equation is to be a function then it can only have one value for each value of x, so that means that the +ive branch of the sqrt is implied.

The x=-3 is actually a valid solution, but only if you allow the sqrt to return both plus and minus results. You can easily verify that x=-3 is indeed a solution so long as you take the sqrt(1) to be minus 1.

Be aware however that once you decide to allow for both +ive and -ive branches of that sqrt then the relationship is multi-valued and at each of the solution points (x-5 and x=-3) you have both x + 1 - 2 sqrt(x+4) = 0 and simultaneously x + 1 - 2 sqrt(x+4) != 0. This is because the relation returns two results for each value of x, one result that is zero and one result that is not.
 
Last edited:
  • #6
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Ok.. I cant really think about it over right now..

but ok what about absolute values

example abs value of (3x-2) = 2 TIMES sqr of (x+8)

How do i do that?
 
  • #7
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How about when you have just abs values on two sides of an equation?

What about when you have sqr roots of x's on both sides?
 
  • #8
Tide
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Solve separately for the two situations where 3x-2 > 0 and 3x-2 < 0 and be sure your solutions comform to those requirements.
 
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  • #9
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Euphoriet said:
Ok.. I cant really think about it over right now..

but ok what about absolute values

example abs value of (3x-2) = 2 TIMES sqr of (x+8)

How do i do that?

abs(x) = sqrt(x^2) therefore abs(3x-2) = sqrt((3x-2)^2), this gives

sqrt((3x-2)^2) = 2*sqrt(x+8)
(3x-2)^2 = 4*(x+8)
This equation you can probably solve, be sure to check the roots ...
 
  • #10
HallsofIvy
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In general, if you square both sides of an equation or multiply both sides of an equation by something involving the unknown x, you may introduce "extraneous roots": values that satisfy the new equation but not the original one.


Obvious example: the equation x= 3 has only 3 as a root but if we multiply both sides by x-2, x(x-2)= 3(x-2) has both 3 and 2 as roots.
 
  • #11
101
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well i still dont fully understand it all. so does anyone have a guide that explains why only one of the answers is correct on the original problem?


Something that talks about having abs value on both sides of the equation woudl also be helpful thanks.
 

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