If roots of the equation ${ax}^2+bx+c=0$ are of the form $$\frac{\alpha}{\alpha-1},\frac{\alpha+1}{\alpha}$$ Then the value of $$({a+b+c})^2$$ I have no clue how to approach this one, any help is appreciated!
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Do you know how to express the product of the roots the equation $$ax^2 + bx + c = 0$$ in terms of $a,b,c$? – quasi Feb 24 '17 at 08:49
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That would be c/a right? – Feb 24 '17 at 08:50
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Yes, that's right. So set c/a = ?? and solve for ?? – quasi Feb 24 '17 at 08:51
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c/a=$\frac{\alpha+1}{\alpha-1}$ – Feb 24 '17 at 08:53
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2Right. Now solve for ??.When that's done, plug ?? back into ?? to get one of the roots of the quadratic in terms of a,b,c. – quasi Feb 24 '17 at 08:54
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Oh so we solve for $\alpha$ and plug it back into the roots, thanks a lot! – Feb 24 '17 at 08:57
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1You only need one of the roots. Take that root, now expressed in terms of a,b,c and plug that back into the quadratic equation. Hopefully, it will reveal the value of (a + b + c)^2. – quasi Feb 24 '17 at 08:59
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Yes that did the trick thanks! – Feb 24 '17 at 08:59
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From the form:
$\frac{1}{x} + x = 2$ => $x^2 + x -2 = 0$, and first equation is $kx^2 + kx - 2k = 0$. So, $(a+b+c)^2=0$
kotomord
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Slick, and much easier than what I was suggesting (but my idea would also work). – quasi Feb 24 '17 at 08:57
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