mkay!

To start, let me show you a bit of "logic":

If A=B then A²=B² (A=B => A²=B²)

eg:

-2=-2

(-2)²= (-2)²

4=4

The opposite not always is true (that's why the 'implies =>' symbol instead of the 'equivalent to <=>'):

A²=B² ≠> A=B

e.g.

A²=B²

4=4

2²=2²

2=2

A=B

A²=B²

16=16

(-4)²=4²

-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*

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Eq.1

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2√(x+4)=3

√(x+4)=3/2 'A=B => A²=B²

x+4= (3/2)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

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Eq.2

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√(x+3)=2√(x)

x+3=2²x

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

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Let try an equation were some solutions really aren't..

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√(x+1)=x-1

x+1=(x-1)²

x+1=x²-2x+1

x=x²-2x

3x-x²=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.

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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.

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&btw...please feel free to register in the forum!

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