Question

Theory of Quadratic Equations, help.

Answer 1

If the roots the equation \(ax^2 + bx + c =0\) are real and of opposite sign then the roots of the equation \(α (x – β)^2 + β (x – α)^2 = 0\) are ?

Answer 2

I expanded the second equation and got :

\(x^2(\alpha + \beta) -x(4\alpha \beta ) + \alpha \beta (\alpha + \beta) = 0 \)

@BAdhi

Answer 3

I suppose alpha and beta are the roots of the first equation, isn't it?

Answer 4

Yes!

Answer 5

My options are :

i) Positive

ii) Real and of opposite sig

iii) imaginary

iv) Negative.

Answer 6

ii) option : sign. (last word)

Answer 7

$$\alpha + \beta=\frac{-b}{a}, \quad \alpha\beta=\frac ca$$
since they are real,$$\Delta>0\implies b^2-4ac>0\implies b^2>4ac$$
since the roots are in opposite sign, $$\alpha\beta<0\implies \frac ca<0$$
therefore,
$$x^2(\alpha + \beta) -x(4\alpha \beta ) + \alpha \beta (\alpha + \beta) = 0\\
x^2\left(\frac{-b}{a}\right)-4x\left(\frac ca\right)+\frac{-b}{a}\frac ca=0\\
(ab)x^2+(4ac)x+(bc)=0$$
$$\Delta_2=(4ac)^2-4(ab)(bc)=4ac(4ac-b^2)$$
with c/a<0 and b^2-4ac>0,
$$ac<0,\;4ac-b^2<0$$
$$\therefore \Delta_2>0$$roots are real
if the roots are $$\gamma,\;\delta$$
$$\gamma\delta=\frac{bc}{ab}=\frac ca<0\implies \text{roots are in opposite sign}$$

Answer 8

Thanks a lot @BAdhi .

Answer 9

you're mostly welcome :)