Question

$\mid x^3-6x^2+11x-6\mid=6$

Answer 1

0, 4

Answer 2

The given equation is $\mid x^3-6x^2+11x-6\mid=6$.
This equation can be split into two separate equations:

1. $x^3-6x^2+11x-6 = 6$
2. $x^3-6x^2+11x-6 = -6$

Let's solve the first equation:
$x^3-6x^2+11x-6 = 6$
$x^3-6x^2+11x-12 = 0$
Let $P(x) = x^3-6x^2+11x-12$. We look for rational roots using the Rational Root Theorem. Possible rational roots are divisors of -12. Let's test integer divisors: $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$.
$P(1) = 1-6+11-12 = -6 \neq 0$
$P(2) = 8-24+22-12 = -6 \neq 0$
$P(3) = 27-54+33-12 = -6 \neq 0$
$P(4) = 4^3 - 6(4^2) + 11(4) - 12 = 64 - 6(16) + 44 - 12 = 64 - 96 + 44 - 12 = 108 - 108 = 0$.
So, $x=4$ is a root.
We can factor $(x-4)$ out of the polynomial $x^3-6x^2+11x-12$ using polynomial division or synthetic division.
Using synthetic division with root 4:

4 | 1  -6  11  -12
  |    4  -8   12
  ----------------
    1  -2   3    0

The quotient is $x^2-2x+3$.
So, $x^3-6x^2+11x-12 = (x-4)(x^2-2x+3) = 0$.
The roots are $x=4$ or $x^2-2x+3=0$.
For the quadratic equation $x^2-2x+3=0$, the discriminant is $\Delta = b^2-4ac = (-2)^2 - 4(1)(3) = 4 - 12 = -8$.
Since the discriminant is negative ($\Delta < 0$), the quadratic equation has no real roots.
Thus, the only real root from the first equation is $x=4$.

Now let's solve the second equation:
$x^3-6x^2+11x-6 = -6$
$x^3-6x^2+11x = 0$
Factor out $x$:
$x(x^2-6x+11) = 0$
The roots are $x=0$ or $x^2-6x+11=0$.
For the quadratic equation $x^2-6x+11=0$, the discriminant is $\Delta = b^2-4ac = (-6)^2 - 4(1)(11) = 36 - 44 = -8$.
Since the discriminant is negative ($\Delta < 0$), the quadratic equation has no real roots.
Thus, the only real root from the second equation is $x=0$.

The real solutions to the original equation $\mid x^3-6x^2+11x-6\mid=6$ are the real roots found from both cases.
The real roots are $x=4$ and $x=0$.

We can check these solutions in the original equation:
For $x=0$: $\mid 0^3 - 6(0)^2 + 11(0) - 6 \mid = \mid -6 \mid = 6$. This is correct.
For $x=4$: $\mid 4^3 - 6(4)^2 + 11(4) - 6 \mid = \mid 64 - 6(16) + 44 - 6 \mid = \mid 64 - 96 + 44 - 6 \mid = \mid 108 - 102 \mid = \mid 6 \mid = 6$. This is correct.

The real solutions are $x=0$ and $x=4$.