REALLY NEED HELP Solving for X with radicals/square roots

  • Context: High School 
  • Thread starter Thread starter Euphoriet
  • Start date Start date
  • Tags Tags
    Roots
Click For Summary

Discussion Overview

The discussion revolves around solving equations involving radicals and absolute values, specifically focusing on the equation x + 1 - 2√(x+4) = 0 and its implications. Participants explore the nature of solutions, the introduction of extraneous roots, and methods for handling absolute value equations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that the equation x + 1 - 2√(x+4) = 0 has a solution of x = 5, but expresses confusion about the validity of x = -3 as a solution.
  • Another participant suggests substituting both values back into the original equation to verify their validity.
  • A different participant notes that squaring both sides of the equation introduces extraneous roots, emphasizing the need for solutions to satisfy the original equation.
  • One participant argues that if the original equation is treated as a function, it should only yield one value for each x, implying that the positive branch of the square root is preferred.
  • Another participant acknowledges that x = -3 can be a valid solution if both positive and negative branches of the square root are considered, but warns that this leads to a multi-valued relationship.
  • Several participants shift the focus to absolute value equations, asking how to solve equations involving absolute values and square roots on both sides.
  • One participant suggests solving separately for cases where the expression inside the absolute value is positive or negative.
  • Another participant provides a method for transforming an absolute value equation into a solvable form, while cautioning about checking the roots for validity.
  • A participant expresses a desire for a guide that explains why only one of the answers is correct in the original problem, particularly in the context of absolute values.

Areas of Agreement / Disagreement

Participants generally agree on the need to check solutions against the original equation to identify extraneous roots. However, there is disagreement regarding the validity of x = -3 as a solution and the implications of treating square roots as multi-valued functions.

Contextual Notes

Participants note the potential for extraneous roots when squaring both sides of equations and the importance of verifying solutions against the original equations. The discussion also highlights the complexity introduced by absolute values and the necessity of considering different cases.

Who May Find This Useful

This discussion may be useful for students grappling with solving equations involving radicals and absolute values, as well as those interested in understanding the implications of extraneous roots in mathematical problem-solving.

Euphoriet
Messages
105
Reaction score
0
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 don't know why but it just does... I can't 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 don't know how its done=/ please 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.
 
Mathematics news on Phys.org
Substitute both values back into the original equation and see what you get!
 
I know I can do that... and I know I can prove that one doesn't work BUT is there any other way of doing it.
 
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!
 
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:
Ok.. I can't 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?
 
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?
 
Solve separately for the two situations where 3x-2 > 0 and 3x-2 < 0 and be sure your solutions comform to those requirements.
 
Last edited:
Euphoriet said:
Ok.. I can't 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
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
well i still don't 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 would also be helpful thanks.
 

Similar threads

  • · Replies 41 ·
2
Replies
41
Views
6K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K