Help with solving radical equations?

Question

Hi, I need help with solving this : 

 

2√x+1=√x^2+4

Answer 1

Hey Alex,

 

The trick is to treat √x as the variable. To do this, we make up a new variable u = √x. Then the equation becomes

2u + 1 = u^2 + 4   =>  u^2 - 2u + 3 = 0  =>  ...  =>  u = ... .

 

You end up with two solutions for u. Then you have to take your two u's and turn them back into x's. It may or may not be true that both (or any) u's lead to good solutions for x -- if u is negative, for example, then there's no real number x such that u=√x is negative, #contradiction. On the other hand, if you have two positive solutions for u, you have two real, positive solutions for x.

 

 

Here's an example that may or may not be similar:

(√x - 4)(√x + 9) = 0

(u - 4)(u + 9) = 0,  u = √x

u = 4 or u = -9

Real square roots can't be negative; throw away u = -9.

u = 4  =>  √x = 4 =>  x = 16

http://www.wolframalpha.com/input/?i=(sqrt(x)+-+4)(sqrt(x)+%2B+9)

 

 

Answer 2

2√x + 1 = √x² + 4

 

√x² - 2√x + 1 = -2

 

√x - 1 = ±i√2

 

√x = 1 ± i√2

 

x = (1±i√2)² = -1 ± 2i√2

 

Notice that this equation has no real roots.

 

We can, however test the two complex roots for whether they work. For complex numbers, √ results in numbers in the 1st and 4th quadrants only.

 

The second to last step above implies that both work.

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