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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 $\frac{c}{a}>0$.

Loo Soo Yong
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  • 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 $x_1, , x_2 > 0$. – Hans Engler Apr 24 '21 at 16:02
  • Have you tried the case when a = b = c = 1? – Mick Apr 25 '21 at 15:39
  • If $2ac-b^2\gt 0$, then the equation does not have any positive solutions. – mathlove Apr 26 '21 at 16:20

1 Answers1

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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).

Bernard
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