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# helppppp

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If a and b are nonzero unequal real numbers and (a-b)/a = b/(a-b) what is the sum of all possible values for a/b?

Mar 28, 2020

#1
+483
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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

Mar 29, 2020
#2
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Very elegant!

AnExtremelyLongName  Mar 29, 2020