form an equation with rational coefficients which has a root 2^(1/2) - 3^(1/3) explainig the steps?

Answer 1

Tricky.

Let x = 2^1/2 - 3^1/3. You want to do enough algebra to make the square and cubes roots go away, until you only have integer (or rational) coefficients.

I would start by isolating the 2^1/2, since it is easier to square a binomial than it is to cube one.

On second thought, that will make a mess of cube roots to deal with. I can handle a mess of square roots more easily, so I change my mind: isolate the 3^1/3 and cube both sides. Move the 3^1/3:

x + 3^1/3 = 2^1/2

Subtract the x.

3^1/3 = 2^1/2 - x.

Now, cube both sides:

(3^1/3)^3 = (2^1/2 - x)^3

3 = (2^1/2 - 3)

I'm not going through all the gory details about cubing the right side. I will just mention that you can use the formula

(A - B)^3 = A^3 - 3*A^2*B + 3*A*B^2 - B^3

to get

3 = 2*2^1/2 - 3*2*x + 3*2^1/2*x^2 - x^3, or
3 = 2*2^1/2 - 6x + 3*2^1/2*x^2 - x^3.

(It's your job to justify that last move).

Now, the only irrational coefficients left are the pair of 2^1/2's. You want to get all of them on one side so you can factor them out. I'm moving the -6x and -x^3 from the right side to the left side to get

x^3 + 6x + 3 = 2*2^1/2 + 3*2^1/2*x^2.

Factor out the 2^1.2 from both sides to get

x^3 + 6x + 3 = 2^1/2 (2 + 3x^2).

Almost home. Square both sides. Three things happen.

1) The left becomes (x^3 + 6x + 3)^2, which I'll leave up to you to FOIL out (it's not hard).
2) The 2^1/2 on the right gets squared and becomes just a 2, and
3) The (2 + 3x^2) on the right becomes (2 + 3x^2)^2, which will also have to be foiled.

If you foil all the mess, you get

FOLED stuff on the left = 2 (FOILED stuff on the right).

Distribute the 2, and move everything to one side (I would move to the left side, so that the highest power, x^6, would be positive), and voila, there's your polynomial.


If you just copy the answer from someone below, then you suck. Answers mean nothing. It's the journey that matters.

Answer 2

this form of quadratic equation is impossible. right this is the evidence: enable the quadratic equation be ax^2 + bx + c = 0, the place a, b and c are unusual. For the quadratic equation to have a rational root, the discriminant might desire to be a applicable sq.. So: b^2 - 4ac = ok^2, the place ok is any integer. As b is unusual, b^2 is likewise unusual. 4ac is often even. So, ok^2 is unusual. Now, enable us to take the finished equation modulo 8: b^2 - 4ac = ok^2 (mod 8) All unusual squares are congruent to a million mod 8. So, b^2 = ok^2 = a million (mod 8). Then we've: a million - 4ac = a million (mod 8) 4ac = 0 (mod 8) this means that 4ac is divisible via 8. enable 4ac = 8x Then, ac = 2x. this means that ac is divisible via 2. this means that the two a or c might desire to be divisible via 2. besides the incontrovertible fact that, a and c are the two unusual, so it incredibly isn't accessible. This leads to a contradiction. for this reason, there exists no quadratic equation with unusual coefficients and a rational root. wish this enables.

Answer 3

I'll do everyone a favor and give cchappa's final answer as:

x^6 - 6x^4 + 6x^3 +12x^2 + 36x + 1 = 0

where, indeed, x = 2^(1/2) - 3^(1/3) is a root.

Repeating what cchappa has said, it works out this way:

3^(1/3) = 2^(1/2) - x
3 = 2*2^(1/2) - 6 x + 3*2^(12)x^2 - x^3
x^3 + 6 x + 3 = 2^(1/2) (2 + 3 x^2)

Square both sides and rearrange.

Answer 4

simple its actuly d difference b/w d square roots of 2 nd 3. so in actual it does nt have any coefficient