Using rational roopts theorem, find the solutions of the equation x^3-6x-4=0.?

Answer 1

x3 - 6x - 4 = 0
x3 - 6x = 4
x ( x2 - 6 ) = 4

there are three roots since the highest power is raised to three.

x = 0
x = + 2.4495 (or +square root of 6)
x = - 2.4495 (or - square root of 6)

Answer 2

TheRationalRootsTheoremStatesThatILoveIV-Rizal

Use this theorem to solve for x³ - 6x - 4 = 0

Joke!

OK I admit I do not remember it but I know that if you have a polynomial equation (of x) of degree n in which p is the coefficient of x^n and q is the constant term, get all integral factors of q/p.

Here, the polynomial equation for x is of degree 3, the coefficient of x³ is 1, and the constant term is -4. Thus, q/p = -4/1 = -4. Hence you will get all integral factors of -4. Thus, they are 4, 2, 1, -1, -2 and -4. Choose any of these and try if it is a solution to the equation. Here I arbitrarily choose 1.
Use 1 as the root and use synthetic division.

_1_| ... 1 .... 0 .... -6 .... -4
_________1___ 1 __ -5
.. ... .....1 .... 1 .... -5 .... -9

Since the remainder is not zero, then 1 is not a factor. Try another one, say -2.

_-2_| ... 1 ... 0 ... -6 ... -4
________ -2 __ 4___ 4
.. ... .... 1 ...-2 .... -2 .... 0

Now that the remainder is zero, -2 is a factor, and the results are coefficient of the other "longer factor".
[x - (-2)][(1) x² + (-2)x + (-2)] = 0
(x + 2)(x² - 2x - 2) = 0

The other factor is now a quadratic, and since the discriminant b² - 4ac is not a perfect square, it cannot be factored anymore. Instead you will use other methods to find the roots of x² - 2x - 2 (like the quadratic formula) and find out that the roots are 2 ± √3.

Therefore, the solutions to the equation x³ - 6x - 4 = 0 are -2, 2 + √3 and 2 - √3.

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Answer 3

exciting. earlier i glance on the hyperlinks some observations: Set y=p/q lowest words, x=r/s lowest words, then p^2s^3 = r^3q^2 - 2q^2s^3 provides s^3 | q^2, q^2 | s^3 and s^3 = q^2 = t^6, giving a diophantine equation (*)............. p^2 = r^3 - 2t^6. For which you have got here upon p=383, r=129, t= 10. The values for the denominators are many times q=t^3 and s=t^2 ; t=a million provides p=5,r=3. At this factor i do no longer comprehend a pair of million

Answer 4

First find leading coefficient (1 is the leading coefficient in front of x^3). Its integer factors are:
q = +/-1

Next find the constant coefficient (-4) and list its integer factors. They are:
p = +/-1, +/-2, +/-4

So the possible rational roots are all forms p/q, or:
+/-1, +/-2, +/-4

Trying all six combinations:
+1 --> 1^3 - 6 - 4 = -9 (bad)
-1 --> -1^3 + 6 - 4 = 1 (bad)
+2 --> 2^3 - 6(2) - 4 = 8 - 12 - 4 = -8 (bad)
-2 --> -2^3 -6(-2) - 4 = -8 +12 - 4 = 0 (GOOD!)
+4 --> 4^3 - 6(4) - 4 = 64 - 24 - 4 = 36 (bad)
-4 --> -4^3 -6(-4) - 4 = -64 + 24 - 4 = -44 (bad)

So x = -2 is one root. That means we can factor out an (x+2) term:
(x+2)(x^2 -2x - 2)

Using the quadratic formula on the second part you get:
x = [ -b +/- sqrt(b^2 - 4ac) ] / 2a

a = 1, b = -2, c = -2
x = [ 2 +/- sqrt(4 + 8) ] / 2
x = 1 +/- sqrt(12)/2
x = 1 +/- sqrt(4)sqrt(3)/2
x = 1 +/- sqrt(3)

So your roots are:
x = -2
x = 1 + sqrt(3) = 2.73205081...
x = 1 - sqrt(3) = -0.73205081...

Answer 5

Ok, I admit I don't remember the rational roots theorem ,but you can also use trial and error method to obtain the first root, then divide the equation by the factor to obtain a quadratic equation.

Trial by error gives me -2 as the first root.

Dividing the equation by x+2 gives

x^2 - 2x - 2.

Now using the quadratic formula, u can find the rest of the solution. I've used it once tonite, so, i don't wanna dabble into the details again.
But the final answers are: 0, (1+1 sqrt 2), (1 - 1sqrt2)

Answer 6

-4 + -6x + x3 = 0

Answer 7

x=2.73

Answer 8

WELL I DONT NO