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 \(b\)?
+118690 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.
+2668 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.
