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!
A question on roots of a Quadratic.
3
$\begingroup$
quadratics
-
0Do you know how to express the product of the roots the equation $$ax^2 + bx + c = 0$$ in terms of $a,b,c$? – 2017-02-24
-
0That would be c/a right? – 2017-02-24
-
0Yes, that's right. So set c/a = ?? and solve for ?? – 2017-02-24
-
0c/a=$\frac{\alpha+1}{\alpha-1}$ – 2017-02-24
-
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. – 2017-02-24
-
0Oh so we solve for $\alpha$ and plug it back into the roots, thanks a lot! – 2017-02-24
-
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. – 2017-02-24
-
0Yes that did the trick thanks! – 2017-02-24
1 Answers
6
From the form:
$\frac{1}{x} + x = 2$ => $x^2 + x -2 = 0$, and first equation is $kx^2 + kx - 2k = 0$. So, $(a+b+c)^2=0$
-
0Slick, and much easier than what I was suggesting (but my idea would also work). – 2017-02-24
-
0Does that mean the other form has no use at all? – 2017-02-26