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# Vieta formula question

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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

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Apr 30, 2022
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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
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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