Hello, I was looking at Riley's solution manual for this specific problem. Along the way, he ended up with a quadratic inequality:

View attachment how.bmp

If you looked at the image, he said it is given that λ is real, but he asserted that λ has no real roots because of the inequality. Doesn't that mean λ is imaginary or complex in some points then, contradicting his first statement? I reckon this has something to do with the properties of a quadratic inequality for complex variables.

pasmith
Homework Helper
Hello, I was looking at Riley's solution manual for this specific problem. Along the way, he ended up with a quadratic inequality:

View attachment 61729

If you looked at the image, he said it is given that λ is real, but he asserted that λ has no real roots because of the inequality. Doesn't that mean λ is imaginary or complex in some points then, contradicting his first statement? I reckon this has something to do with the properties of a quadratic inequality for complex variables.

$\lambda$ doesn't have roots; it's an arbitrary real number. The quadratic $P: z \mapsto az^2 + bz + c$ has roots, which are those $z \in \mathbb{C}$ for which $P(z) = 0$. The point is that if $P(\lambda) > 0$ for all real $\lambda$ then $P$ has no real roots, because if $z$ is real then $P(z) \neq 0$ and $z$ cannot be a root of $P$.

That's cool, I get it now. Then that means the roots are either purely imaginary or complex (but not purely real) right? Then why is it required that the discriminant be less than zero? Is it because it will make sure that part of the solution will have an imaginary part?

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pasmith
Homework Helper
That's cool, I get it now. Then that means the roots are either purely imaginary or complex (but not purely real) right? Then why is it required that the discriminant be less than zero? Is it because it will make sure that part of the solution will have an imaginary part?

Yes. A quadratic with real coefficients has no real roots if and only if the discriminant is negative.