Question: To show that the equation $4x^3 = 9x^2 - 10x - 8$ has a negative real solution (i.e. a solution $<0$).
My Attempt:
Given equation: $4x^3 = 9x^2 - 10x - 8$
Let $f(x) = 4x^3 - 9x^2 + 10x + 8$
We have:
$f(0) = 4(0)^3 - 9(0)^2 + 10(0) + 8 = 8$
$f(-1) = 4(-1)^3 - 9(-1)^2 + 10(-1) + 8 = -4 - 9 - 10 + 8 = -15$
Therefore, $f(-1) < 0 < f(0)$, which means that $f(x)$ has a real root between $-1$ and $0$ by the Intermediate Value Theorem. Since this root is negative, it is a negative real solution of the given equation.
Therefore, the given equation has a negative real solution, as required.
Am I right ?
That is right. Moreover, it is the only real root of the equation, as there can't be odd number of complex roots.