Hey guest! let's first write out what we're given:
\(\frac{a-b}a = \frac b{a-b}\)
Cross multiplying, we can write:
\(ab = (a-b)^2\)
Expand this out, and we get:
\(ab = a^2-2ab + b^2\)
Subtracting ab on both sides, we then get:
\(a^2-3ab + b^2 = 0\)
Now, we divide by b2(and you'll see why pretty soon). This gives us:
\((\frac ab)^2 -3(\frac a b) + 1 = 0\)
Let's then name \(\frac a b\)as x(Do you see why? Because it makes for easy substitution for a quadratic. If we define it as x, if we solve for the values of x, we solve for the values of \(\frac a b\))
Substituting into our quadratic, we get:
\(x^2-3x+1 = 0\)
Clearly, this isn't factorable with integer roots, so we use the quadratic formula. This then gives us:
\((-(-3)\pm\sqrt{(-3)^2-4*1*1})/2 = (3\pm\sqrt{9-4})/2 = (3\pm\sqrt{5})/2\) as our two possible values of a/b. However, we're not done yet. The question asks us for the sum of all possibles values for \(\frac ab\). We can see that our two roots are the conjugates of each other, meaning that their irrational part cancels out. This then leaves us with:
\(\frac {3+ \sqrt5} {2} + \frac{3-\sqrt5} {2} = \frac 6 2 = 3\) as our final answer
