Question

The equation \( 6x^4 - 5x^3 + 13x^2 - 5x + 6 = 0 \) will have:

• A. only real roots
• B. only complex roots
• C. two real and two complex roots
• D. two real and two purely imaginary roots

Hint:

When no rational or real roots exist in a real-coefficient polynomial, and the degree is even, then all roots must be complex and appear in conjugate pairs.

Answer 1

The Correct Option is B

Step 1: Consider the polynomial: \[ f(x) = 6x^4 - 5x^3 + 13x^2 - 5x + 6 \] Step 2: Apply Descartes' Rule of Signs for real roots:
- For \( f(x) \): sign changes = \( -, +, -, + \Rightarrow 3 \) changes → possibly 3 or 1 positive real roots.
- For \( f(-x) \): \[ f(-x) = 6x^4 + 5x^3 + 13x^2 + 5x + 6 \Rightarrow \text{All terms are positive} \Rightarrow 0 \text{ sign changes} \Rightarrow 0 \text{ negative real roots.} \]
Conclusion: At most 3 positive real roots, but we verify by trying rational roots.
Step 3: Use Rational Root Theorem:
Possible rational roots: \( \pm1, \pm2, \pm3, \pm6, \pm\frac{1}{2}, \pm\frac{3}{2}, \pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{1}{6} \)
Try testing some values:
\[ f(1) = 6 - 5 + 13 - 5 + 6 = 15 \neq 0 \quad \text{(not a root)} \]
\[ f(-1) = 6 + 5 + 13 + 5 + 6 = 35 \neq 0 \quad \text{(not a root)} \]
\[ f(2) = 6(16) - 5(8) + 13(4) - 5(2) + 6 = 96 - 40 + 52 - 10 + 6 = 104 \neq 0 \quad \text{(not a root)} \]
No rational root exists.
Step 4: Use the Discriminant or Factorization approach:
No rational roots implies no real linear factors. All roots are non-real complex or irrational.
Also, since the polynomial has real coefficients, non-real complex roots must occur in conjugate pairs.
Step 5: Since no real roots exist (from above), and all 4 roots must be complex (and conjugate pairs), the equation has:
\[ \boxed{\text{Only complex roots}} \]