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does √x-y = √x-√y

Guest Jan 12, 2016

Best Answer 

 #2
avatar+78719 
+5

√x-y = √x-√y   (???)    square both sides

 

x - y = x - 2√[x *y]  + y    subtract x from both sides  and rearrange

 

2√[x *y]  = 2y        divide both sides by 2

 

√[x * y]  = y

 

√x * √y  = y

 

√x * √y  - y  =  0        factor

 

√y [ √x  - √y]  = 0

 

So either    √y  = 0      or    √x  = √y

 

So....the original equation is true only when x = y   or    y = 0

 

 

cool cool cool

CPhill  Jan 12, 2016
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3+0 Answers

 #1
avatar+1828 
+5

NOOOOOO 

You can't 

it works for multiplcation or division 

xvxvxv  Jan 12, 2016
 #2
avatar+78719 
+5
Best Answer

√x-y = √x-√y   (???)    square both sides

 

x - y = x - 2√[x *y]  + y    subtract x from both sides  and rearrange

 

2√[x *y]  = 2y        divide both sides by 2

 

√[x * y]  = y

 

√x * √y  = y

 

√x * √y  - y  =  0        factor

 

√y [ √x  - √y]  = 0

 

So either    √y  = 0      or    √x  = √y

 

So....the original equation is true only when x = y   or    y = 0

 

 

cool cool cool

CPhill  Jan 12, 2016
 #3
avatar+1794 
+5

If you mean \(\sqrt{x-y}\), you cannot seperate \(x\) and \(y\) into \(\sqrt{x}-\sqrt{y}\). If \(x\) and \(y\) was \(\sqrt{x\times y}\), then you can seperate \(x\) and \(y\) into \(\sqrt{x}\times\sqrt{y}\).  If \(x\) and \(y\) was \(\sqrt\frac{{x}}{y}\), then you an seperate \(x\) and \(y\) into \(\frac{{\sqrt x}}{\sqrt y}\).  If you mean \(\sqrt{x}-y\), you cannot add a \(\sqrt{}\) to y, that would change the problem entirely.

 

Example:

 

Let \(x=3\) and let \(y=2\)

 

\(\sqrt{x-y}\)

 

\(\sqrt{3-2}\)

 

\(\sqrt{1}\)

 

\(1\)

 

The answer is \(1\)

 

Now lets do \(\sqrt{x}-\sqrt{y}\)

 

Let \(x=3\) and let \(y=2\)

 

\(\sqrt{x}-\sqrt{y}\)

 

\(\sqrt{3}-\sqrt{2}\)

 

\(1.7320508075688773...-1.414213562373095...\)

 

\(0.3178372451957823...\)

 

The answer is \(0.3178372451957823...\)

 

\(1\) and \(0.3178372451957823...\) are not the same answer.

 

Now lets do \(\sqrt{x\times y}\)

 

Let \(x=3\) and let \(y=2\)

 

\(\sqrt{x\times y}\)

 

\(\sqrt{3\times 2}\)

 

\(\sqrt{6}\)

 

\(2.4494897427831781...\)

 

The answer is \(2.4494897427831781...\)

 

Now lets do \(\sqrt{x}\times\sqrt{y}\)

 

Let \(x=3\) and let \(y=2\)

 

\(\sqrt{x}\times\sqrt{y}\)

 

\(\sqrt{3}\times\sqrt{2}\)

 

\(1.7320508075688773... \times 1.414213562373095...\)

 

\(2.4494897427831781...\)

 

The answer is \(2.4494897427831781...\)

 

\(2.4494897427831781...\) and \(2.4494897427831781...\) are the same answer.

 

Now lets do \(\sqrt\frac{{x}}{y}\)

 

Let \(x=3\) and let \(y=2\)

 

\(\sqrt\frac{{x}}{y}\)

 

\(\sqrt\frac{{3}}{2}\)

 

\(1.224744871391589...\)

 

The answer is \(1.224744871391589...\)

 

Now lets do \(\frac{{\sqrt x}}{\sqrt y}\)

 

Let \(x=3\) and let \(y=2\)

 

\(\frac{{\sqrt x}}{\sqrt y}\)

 

\(\frac{{\sqrt 3}}{\sqrt 2}\)

 

\(\frac{1.7320508075688773...}{1.414213562373095...}\)

 

\(1.224744871391589...\)

 

The answer is \(1.224744871391589...\)

 

\(1.224744871391589...\) and \(1.224744871391589...\) is the same answer

 

Now lets do \(\sqrt{x}-y\)

 

Let \(x=3\) and let \(y=2\)

 

\(\sqrt{x}-y\)

 

\(\sqrt{3}-2\)

 

\(1.7320508075688773...-2\)

 

\(-0.2679491924311227...\)

 

The answer is \(-0.2679491924311227...\)

 

Now let do \(\sqrt{x}-\sqrt y\)

 

Let \(x=3\) and let \(y=2\)

 

\(\sqrt{x}-\sqrt y\)

 

\(\sqrt{3}-\sqrt 2\)

 

\(1.7320508075688773...-1.414213562373095...\)

 

\(0.3178372451957823...\)

 

The answer is \(0.3178372451957823...\)

 

\(-0.2679491924311227...\) and \(0.3178372451957823...\) are not the same answer

 

Hope this helps

gibsonj338  Jan 12, 2016

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