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)