Solve the quintic equation 40x^5 - 51x^4 + 10 = 0.

Step-by-step solution

1. 1

   Rule out rational roots. Candidates are $\pm\tfrac{d}{e}$ with $d \mid 10$ and $e \mid 40$. Spot checks: $f(1) = -1$, $f(-1) = -81$, $f\!\left(\tfrac12\right) = \tfrac{129}{16} = 8.0625$, $f\!\left(-\tfrac12\right) = \tfrac{89}{16} = 5.5625$, $f\!\left(\tfrac54\right) = 7.5586$. None vanish, and by Abel–Ruffini a general quintic has no radical solution, so the roots must be located numerically.

2. 2

   Bracket the real roots by sign changes. Tabulating $f(x) = 40x^5 - 51x^4 + 10$:

   $f(-1) = -81, \quad f(-0.7) = -8.968, \quad f(-0.5) = 5.563, \quad f(0) = 10, \quad f(0.9) = 0.159, \quad f(1) = -1, \quad f(1.1) = -0.249, \quad f(1.2) = 3.779$

   Each sign change traps a root: one in $(-0.7,\,-0.5)$, one in $(0.9,\,1)$, and one in $(1.1,\,1.2)$. Since $f \to -\infty$ as $x \to -\infty$ and $f \to +\infty$ as $x \to +\infty$, these three are all the real roots.

3. 3

   Refine each bracket. Bisection or Newton on each interval converges to

   $x_1 \approx -0.603957, \qquad x_2 \approx 0.909281, \qquad x_3 \approx 1.110778$

4. 4

   Recover the remaining conjugate pair. A quintic has five roots. Deflating by the three real ones leaves a quadratic whose roots are

   $x \approx -0.070551 \pm 0.636284\,i$

5. 5

   Verify with the coefficient relations. The sum of all five roots must be $\tfrac{51}{40} = 1.275$: $-0.603957 + 0.909281 + 1.110778 + 2(-0.070551) = 1.275000$. The product must be $-\tfrac{10}{40} = -0.25$ (degree 5, so the sign flips): $(-0.603957)(0.909281)(1.110778)\left(0.070551^2 + 0.636284^2\right) = -0.250000$. Direct substitution gives $|f(x_i)| < 10^{-12}$ for each real root.

6. 6

   A warning about widely circulated wrong values. The triple $-0.6487,\ 0.7606,\ 1.2264$ is sometimes quoted for this equation, but $f(-0.6487) = -3.63$, $f(0.7606) = 3.11$ and $f(1.2264) = 5.60$ — none of them is a root.