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**Help****Guest**

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?

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,619

1. Looks right!

2.

√(x+3)=2 √x

√(x+3)²=2² √x²

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

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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

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

-----------------------------------------------------------------------

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

-----------------------------------------------------------------------

√(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

-----------------------------------------------------------------------

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

-----------------------------------------------------------------------

√(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.

------------------------------------------------------

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

*Last edited by kylekatarn (2005-08-02 01:17:53)*

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**Help****Guest**

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

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,619

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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