Prove that two coefficients of a quadratic function are always positiveLet f(x) be a quadratic function of the form x2+px+q such that the equation f(f(x))=0 has two equal real roots. Show that p≥0 and q≥0.

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Prove that two coefficients of a quadratic function are always positiveLet f(x) be a quadratic function of the form x2+px+q such that the equation f(f(x))=0 has two equal real roots. Show that p≥0 and q≥0.

davane6a7a · Accepted Answer

Since f(f(x))=0 does have a solution, f(x) can be factored, into say f(x)=(x−α)(x−β). Here, we have α+β=−p,αβ=q.If α=β, you can show α and β must be 0. In this case, we have p=q=0.Now, assume α≠β. Then, f(f(x))=(f(x)−α)(f(x)−β)=0 is equivalent to f(x)−α=0 or f(x)−β=0. Since both of these equations cannot be true at the same time (since α≠β) and f(f(x))=0 has only one solution, it must be the case that one of the two equations has only one root (i.e., a double root), and the other no roots.Without loss of generality, we can assume f(x)−α=0 has a double root, and f(x)−β=0 has no roots. Looking at the discriminant, we get (−α−β)2−4(αβ−α)=0 and (−α−β)2−4(αβ−β)&amp;lt;0.Now, we have (−α−β)2−4(αβ−α)=(α−β)2+4α=0, so α≤0. Similarly, we have β&amp;lt;0. Hence, p=−α−β≥0, and q=αβ≥0

Prove that two coefficients of a quadratic function

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