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The sum \(a+b\), the product \(a⋅b\) and the difference of squares \(a^2-b^2\) of two positive numbers \(a\) and \(b\) is the same non-zero number. What is \(b\)

 Apr 30, 2022
 #1
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b = (1 + sqrt(3))/2

 Apr 30, 2022
 #2
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Hmm? Are you sure about that answer?

 Apr 30, 2022
 #3
avatar+117104 
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Tet all those exparessions = T

 

\(a^2-b^2=(a-b)(a+b)\\ T=(a-b)T\\ so\\ a-b=1\\so\\a=1+b\\~\\ a+b=ab\\ 1+b+b=b(1+b)\\ 1+2b=b+b^2\\ b^2-b-1=0\\ b=\frac{-1\pm\sqrt{1+4}}{2}\\ b=\frac{-1+\sqrt{5}}{2}\\ b=\frac{\sqrt{5}-1}{2}\qquad a=\frac{\sqrt{5}+1}{2}\\ \text{But a and b are interchangeable so}\\ b=\frac{\sqrt{5}\pm1}{2}\ \)

 

 

You need to check it.

 Apr 30, 2022
 #4
avatar+1351 
+1

We have the following system: 

 

\(a+b=x\)                          (1)

\(ab=x\)                               (2)

\((a-b)(a+b)=x\)         (3)

 

Substituting (1) into (3), we get: \((a-b)x=x\)

 

Dividing both sides by x, we find \(a-b=1\)

 

Solving for \(a\) in terms of b, we find: \(a=b+1\)

 

Substituting this into (2), we get: \(b(b+1)=x\)

 

Simplifying, we find: \(b^2+b=x\)

 

Substituting this in (3), we get: \(a^2-b^2=b^2+b\)

 

We also know that \(a = b+1\)

 

Solving this, we find: \({a={ {3 + \sqrt5} \over 2}}\) and \({b={{1 + \sqrt5} \over 2}}\)

 

However, these are interchangeable, so b can be either of these. 

BuilderBoi  Apr 30, 2022

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