Find the three solutions for cubic equation x^3-7x+6=0?

Question

solve this cubic equation and provide a solution

Answer 1

The solutions for the cubic equation x^3 - 7x + 6 = 0 can be found using various methods, such as factoring, the rational root theorem, or numerical methods. However, providing the solutions within the given word limit is not feasible. I recommend using a mathematical software or calculator to obtain the precise solutions, which are approximately x = -1, x = 2, and x = 3.

Answer 2

There is no real root. There is imaginary roots. So this question has no real answer. There is no need even thow here is imaginary roots x=1 and x=-1.so that's it.

Answer 3

To find the solutions for the cubic equation x^3 - 7x + 6 = 0, you can use various methods, such as the Rational Root Theorem, synthetic division, or numerical methods like Newton-Raphson. Here, I'll use the Rational Root Theorem to find potential rational solutions and then use synthetic division to confirm them:

The Rational Root Theorem states that if a rational number p/q is a root of the polynomial equation, then p is a factor of the constant term (here, 6), and q is a factor of the leading coefficient (here, 1).

The factors of 6 are ±1, ±2, ±3, and ±6.

The factors of 1 are ±1.

Now, we'll try these potential rational solutions one by one:

1. x = 1

   - Plug x = 1 into the equation: (1^3) - 7(1) + 6 = 1 - 7 + 6 = 0.

   - x = 1 is a root of the equation.

Now that we have one solution, we can use synthetic division to find the other solutions:

Perform synthetic division with (x - 1) to factor out (x - 1) from the equation:

```

      1 | 1 0 -7 6

         | - 1 -6

         |_______________

           1 1 -6 0

```

The resulting quadratic equation is x^2 + x - 6 = 0, which can be factored as (x + 3)(x - 2).

2. Using x + 3 = 0, we get x = -3.

3. Using x - 2 = 0, we get x = 2.

So, the three solutions to the cubic equation x^3 - 7x + 6 = 0 are:

1. x = 1

2. x = -3

3. x = 2.

Answer 4

To determine the solutions for the cubic equation x^3 - 7x + 6 = 0, we can use several methods. One approach is to use the Rational Root Theorem, which states that if a polynomial equation has a rational root p/q, then p must be a factor of the constant term (in this case, 6) and q must be a factor of the leading coefficient (which is 1).

In this case, the possible rational roots are ±1, ±2, ±3, and ±6. By testing these roots using synthetic division, we find that x = 1 is one solution:

1 | 1 - 7 0 6

    1 - 6 -6

__________________

   1 - 6 -6 0

The resulting equation is x^2 - 6x - 6 = 0. We can then solve this quadratic equation using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 1, b = -6, and c = -6. Substituting these values into the quadratic formula:

x = (-(-6) ± √((-6)^2 - 4(1)(-6))) / (2(1))

  = (6 ± √(36 + 24)) / 2

  = (6 ± √60) / 2

  = (6 ± 2√15) / 2

  = 3 ± √15

Therefore, the two additional solutions to the cubic equation x^3 - 7x + 6 = 0 are x = 3 + √15 and x = 3 - √15.

In summary, the three solutions for the cubic equation x^3 - 7x + 6 = 0 are = 1, x = 3 + √15, and x = 3 -15.