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How should I go around on proving the quadratic equation $$a^2 x^2 +(2ac-b^2)x+c^2=0$$ having real and positive solutions?

I tried to use the fact that if a quadratic equation has real and positive solutions, then the discriminant is greater or equal to 0, and that $\frac{b}{a}<0$ and $\frac{c}{a}>0$. But I kind of stuck after proving that $\frac{c^2}{a^2}>0$ and that $\frac{c}{a}>0$.

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  • $\begingroup$ In your case, you need the coefficient of $x$ to be $<0$, in addition to the discriminant being positive. That is because the quadratic expression can be written as $a^2(x-x_1)(x-x_2) = a^2x^2 - a^2(x_1 + x_2)x + a^2 x_1x_2$ with $x_1, \, x_2 > 0$. $\endgroup$ Commented Apr 24, 2021 at 16:02
  • $\begingroup$ Have you tried the case when a = b = c = 1? $\endgroup$
    – Mick
    Commented Apr 25, 2021 at 15:39
  • $\begingroup$ If $2ac-b^2\gt 0$, then the equation does not have any positive solutions. $\endgroup$
    – mathlove
    Commented Apr 26, 2021 at 16:20

2 Answers 2

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Hint:

This is a standard high school exercise. You indeed have to prove that $$\Delta=(2ac-b^2)^2-4a^2c^2=b^2(b^2-4ac)$$ is positive for the existence of real roots, that the product of these roots is positive (nonzero roots with the same sign) and their sum is also positive (the common sign is also the sign of the sum).

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Firstly, note that since $a^{2}>0$ and $c^{2}>0$ if the quadratic has any roots they will either both be positive or both be negative.

You could use Descarte's rule of signs to then argue that $2ac-b^{2}<0$ for two positive roots but unless you know more about $a,b,c$ this merely gives conditions for the solution rather than the solution itself.

It seems like something is therefore missing from the question.

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