Polynomials
Irrational roots of cubic
MJAT_TS8_P1
Grade 12
Question:
The number of irrational roots of the equation $x^3-6x^2+12x-1=0$ is/are:
Step-by-Step Solution
Key Concept: Rewrite: $(x-2)^3 = -7$. So $x-2=(-7)^{1/3}$: one real cube root ($-7^{1/3}$, real and irrational) and two complex cube roots ($-7^{1/3}\omega$ and $-7^{1/3}\omega^2$). Only one real irrational root.
Only $\mathbf{1}$ irrational root. Answer: B.
Correct Answer: B