Question

If `3x^2-4x+1` has zeros `alpha` and `beta`, then what quadratic has zeros `alpha^2/beta` and `beta^2/alpha` ?

Answer 1

Find `alpha` and `beta` first.

Explanation:

`3x^2 - 4x + 1 = 0`
The left side factors, so that we have
`(3x - 1)(x - 1) = 0`.
Without loss of generality, the roots are `alpha = 1` and `beta = 1/3`.

`alpha^2/beta = 1^2/(1/3) = 3` and `(1/3)^2/1= 1/9`.

A polynomial with rational coefficients having these roots is
`f(x) = (x - 3)(x - 1/9)`
If we desire integer coefficients, multiply by 9 to obtain:
`g(x) = 9(x - 3)(x - 1/9) = (x - 3)(9x - 1)`
We may multiply this out if we wish:
`g(x) = 9x^2 - 28x + 3`

NOTE: More generally, we might write
`f(x) = (x - alpha^2/beta)(x - beta^2/alpha)`
`= x^2 - ((alpha^3 + beta^3)/(alphabeta))x + alphabeta`

Answer 2

`9x^2-28x+3`

Explanation:

Note that:

`(x-alpha)(x-beta) = x^2-(alpha+beta)x+alpha beta`

and:

`(x-alpha^2/beta)(x-beta^2/alpha) = x^2-(alpha^2/beta+beta^2/alpha)x+(alpha^2/beta)(beta^2/alpha)`

`color(white)((x-alpha^2/beta)(x-beta^2/alpha)) = x^2-(alpha^3+beta^3)/(alpha beta) x + alpha beta`

`color(white)((x-alpha^2/beta)(x-beta^2/alpha)) = x^2-((alpha+beta)^3-3alpha beta(alpha+beta))/(alpha beta) x + alpha beta`

In our example, dividing `3x^2-4x+1` by `3` we have:

`{ (alpha+beta = 4/3), (alpha beta=1/3) :}`

So:

`((alpha+beta)^3-3alpha beta(alpha+beta))/(alpha beta) = ((4/3)^3-3(1/3)(4/3))/(1/3) = (64/27-4/3)/(1/3) = 28/9`

So the desired polynomial can be written:

`x^2-28/9x+1/3`

Multiply through by `9` to get integer coefficients:

`9x^2-28x+3`

Answer 3

Proposed solution below;

Explanation:

`3x²-4x+1`

Note: `a` is alpha, `b` is beta

`a + b = 4/3`

`ab =1/3`

To form an equation we find the sum and products of the roots..

For Sum

`(a²)/b + (b²)/a = (a^3 + b^3)/(ab)`

But; `a^3 + b^3 = (a+b)³-3ab(a+b)`

Therefore;

`((a+b)³-3ab(a+b))/(ab)`

Hence we substitute the values..

`((4/3)³-3(1/3)(4/3))/(1/3)`

`((64/27)-cancel3(1/cancel3)(4/3))/(1/3)`

`(64/27 - 4/3)/(1/3)`

`((64 - 36)/27)/(1/3)`

`(28/27)/(1/3)`

`(28/27) div (1/3)`

`(28/27) xx (3/1)`

`(28/cancel27_9) xx (cancel3/1)`

`28/9`

Hence, the sum is `28/9`

For Products

`((a²)/b)((b²)/a)`

`((ab)²)/(ab)`

`(1/3)^2 div 1/3`

`1/9 div 1/3`

`1/9 xx 3/1`

`1/cancel9_3 xx cancel3/1`

`1/3 xx 1/1`

`1/3`

Hence, the product is `1/3`

`x²-(a+b)x+ab`

`x²-(28/9)x+1/3`

`9x²-28x+3`

Multiplying through by `9`

Hope this helps!