show that the equation x^4-4x^3-2x^2+12x+12=0 has no real solutions?

Answer 1

We know the function (call it f(x)) is positive for large positve or large negative values. One way to prove it has no real solution is to show f(x) can never have a negative value. To prove this, it is sufficient to test the points where the slope is 0, since those are the points where f(x) hits a minimum or maximum.

To find the minimum points, take the first derivaitve: f'(x) = 4x^3 - 12x^2 - 4x + 12. This can be factored as (x - 3)(x - 1)(x + 1).

So, f(3) is a minimum, f(1) is a maxiimum, and f(-1) is a minimum. The rest is just computation - calculate f(3) and f(-1) to see if either of them are negative:

f(3) = 3

f(-1) = 3

In conclusion, this function never hits the x-axis.

Answer 2

Solve with the quadratic formula
Quadratic Equation



Result
To make the calculations easier, we can use the fact that the coefficients 2, 12 and 12 have a common factor of 2.

Dividing both sides by 2 then gives the simpler but equivalent equation :

x2 + 6 x + 6 = 0

The quadratic equation

a x2 + b x + c = 0

can always be solved by substituting the coefficients a , b and c into the quadratic formula :


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-b ± b2 - 4 a c
x =
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2 a


To make the following working a bit clearer, we can rewrite the equation

x2 + 6 x + 6 = 0

as

1 x2 + 6 x + 6 = 0.

Comparing this with

a x2 + b x + c = 0

shows that a = 1 , b = 6 and c = 6.

Substituting these values into the quadratic formula then gives


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-6 ± 62 - 4 x 1 x 6
x =
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2 x 1



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-6 ± 36 - 24
x =
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2



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-6 ± 12
x =
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--------------------------------------------------------------------------------

2



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-6 ± 2 3
x =
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2



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2 -3 ± 3
x =
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2




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x = -3 ± 3



Notice that we used this result in the above calculation :


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12
=
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4 x 3
=
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4
x
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3
= 2
--------------------------------------------------------------------------------

3



So there are two real solutions :



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x = -3 ± 3





These solutions are approximately

x = -1.26795 , x = -4.73205

Answer 3

The x^4 dominates when x is very large or small, making the function positive.

Differentiate the function to find the local max and min. This would give you a polinomial whose main term is x^3 guaranteeing at least one real root. Find one by numerical methods then calculate the others using the quadratic formula.

Show that at the local minima the function is positive. This should suffice.

Answer 4

The function has no 0-crossings. Its 2 minima are (-1,3) and (3,3).

Answer 5

using synthetic division, u can solve this easily. Knowing that its index is 4, u know that its (X+A)(X+B)(X+c)(X+D), use i to try it using synthetic division and u will get solution, U can also use graphic calculator, plot it

Answer 6

Sorry, i thought about the question.. but didnt know. good luck anyways