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#1 2005-08-02 22:05:00

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Solving Radical equations

Is this solution correct?

1.
2[SQRT(x+4)]=3
2[SQRT(x+4)]^2=3^2
4x+16=9
4x=9-16
4x=-7
x=-7/4

2.
[SQRT(x+3)]=2[SQRT(x)]
[SQRT(x+3)]^2=2[SQRT(x)]^2
x+9=4x I'm not sure what to do here?

#2 2005-08-02 22:36:54

MathsIsFun
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Re: Solving Radical equations

1. Looks right!

2.
√(x+3)=2 √x
√(x+3)=2 √x
x+3 = 4x    [Note: the square root and square cancel each other out, just leaving (x+3)]

Now, subtract 4x from both sides:  x+3 - 4x = 4x - 4x
Simplify:  -3x + 3 = 0
So, x must be 1

(Or, you could be formal, and follow the steps:
Subtract 3 from both sides: -3x = -3
divide by -3: x = 1)


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

#3 2005-08-02 23:16:41

kylekatarn
Power Member

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Re: Solving Radical equations

mkay!

To start, let me show you a bit of "logic":
If A=B then A=B (A=B => A=B)
eg:
-2=-2
(-2)= (-2)
4=4

The opposite not always is true (that's why the 'implies =>' symbol instead of the 'equivalent to <=>'):
A=B ≠> A=B
e.g.

A=B
4=4
2=2
2=2
A=B

A=B
16=16
(-4)=4
-4≠4
A≠B

This type of equations do not always return valid roots!
So how to proceed in order to ensure that all solutions are valid?

It's easy but very important!
-> When you'r done solving the equation, you must plug in ALL solutions into the original equation and verify if you get an equality (1=1,  x=x,... something=something)
If that does not happen you must discard that solution.

*Lets check your equations and then tryout a more complex ones*
-----------------------------------------------------------------------
Eq.1
-----------------------------------------------------------------------
2√(x+4)=3
√(x+4)=3/2     'A=B => A=B
x+4= (3/2)
x=9/4-4=-7/4

verification:
2√(-7/4+4)=3
2*1.5=3
3=3         'TRUE
the first eq. has solution x=-7/4

-----------------------------------------------------------------------
Eq.2
-----------------------------------------------------------------------
√(x+3)=2√(x)
x+3=2x
x+3=4x
3=4x-x
3=3x
x=1

verification:
√(1+3)=2√(1)
√(4)=2
2=2             'TRUE
the second eq. has solution x=1

-----------------------------------------------------------------------
Let try an equation were some solutions really aren't..
-----------------------------------------------------------------------
√(x+1)=x-1
x+1=(x-1)
x+1=x-2x+1
x=x-2x
3x-x=0
(3-x)x=0
x=3 or x=0

Lets verify both (supposed) solutions
x=3
√(3+1)=3-1
√(4)=2
2=2     'TRUE
great! x=3 is a solution!

Now for x=0
√(0+1)=0-1
√(1)=-1
1=-1     FALSE!!!
X=0 ISN'T A ROOT(SOLUTION) OF √(x+1)=x-1

See how important it is to verify your solutions?
I hope it was all very clear.

-----------------------------------------------------------------------
All the above stuff has to do also with the domains os the √() operation in IR.
√[u(x)]=v(x) => u(x)≥0 ...we cant have sqrt of neg. numbers
You can use domains in order to validate solutions but I think that substituition is the faster & simplier method. You choose.
------------------------------------------------------
&btw...please feel free to register in the forum!

Last edited by kylekatarn (2005-08-02 23:17:53)

#4 2005-08-03 20:36:40

Help
Guest

Re: Solving Radical equations

Thanks to both of you! I wasn't sure how to check for valid solutions, but now I do!smile I'll register once I come up with a username. I'll need all the math help I can get.^^;

#5 2005-08-03 23:11:45

MathsIsFun
Administrator

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Re: Solving Radical equations

We look forward to more "radical questions" ...

Don't forget that a square root is just a half power:

√x = x^(1/2)


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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