If alpha and beta are the zeroes of the polynomial f (x) = 5y square -7y + 1. Find another polynomial whose zeroes are 2 alpha / beta and 2beta / alpha

Whose zeroes ar 1/ 2alpha + beta and 1/2beta + alpha

Let alpha = a

Let beta = b
Because it's easier to type a and b, than copy-pasting alpha and beta symbols.
Anyway, recall that for a quadratic equation in the form
y = ax^2 + bx + c
The sum and product of the roots can be determined as such:
y = x^2 - (sum)x + (product)
Of course, the numerical coefficient of x^2 must be 1.

Applying this on the given function,
5y^2 - 7y + 1
(1/5) (y^2 - (7/5)y + 1/5)
Therefore, sum of the roots (a + b) and product of the roots (ab), respectively, are:
a + b = 7/5
ab = 1/5

Now, we are asked to find a polynomial with roots 2a/b and 2b/a. What we'll do is get their sum and product:
Product:
2a/b * 2b/a
4ab / ab
= 4

Sum:
2a/b + 2b/a
(2a^2 + 2b^2) / ab
2(a^2 + b^2) / ab
Complete the square inside the parenthesis by adding 2ab, but at the same time subtracting 2(2ab) outside the parenthesis to counter its effect:
[ 2(a^2 + 2ab + b^2) - 4ab ] / ab
[ 2(a + b)^2 - 4ab ] / ab
note that, we have values for a + b and ab from above.
[ 2(7/5)^2 - 4(1/5) ] / (1/5)
= 78/5

Therefore, the polynomial is,
y^2 - (sum)y + (product)
y^2 - (78/5)y + 4

Now try the other problem by using this method.
hope this helps~ `u`

To find another polynomial whose zeroes are 2α/β and 2β/α, we can use the fact that the sum and product of the zeroes of a polynomial are related to its coefficients.

Let's start by finding the sum and product of the zeroes of f(x) = 5x^2 - 7x + 1.

The sum of the zeroes (α + β) can be found using the formula: sum = -b/a, where the polynomial is in the form ax^2 + bx + c. In this case, a = 5 and b = -7.

Sum of zeroes (α + β) = -(-7)/5 = 7/5

The product of the zeroes (α * β) can be found using the formula: product = c/a, where c is the constant term and a is the coefficient of x^2. In this case, a = 5 and c = 1.

Product of zeroes (α * β) = 1/5

Now, we can find the zeroes of the new polynomial whose zeroes are 2α/β and 2β/α.

Let's call the new polynomial g(x).

The sum of the new zeroes can be calculated as follows:
sum of new zeroes = (2α/β) + (2β/α) = 2(α/β + β/α) = 2(α^2 + β^2)/(αβ)

Using the identity (α + β)^2 - 2αβ = α^2 + β^2, we can rewrite the above expression as:
sum of new zeroes = 2((α + β)^2 - 2αβ)/(αβ) = 2((7/5)^2 - 2(1/5))/(1/5) = 10(7^2 - 2)/(1/5)
= 350 - 100
= 250

Therefore, the sum of the new zeroes is 250.

Similarly, the product of the new zeroes can be calculated as follows:
product of new zeroes = (2α/β) * (2β/α) = 4(αβ)/(αβ) = 4

Therefore, the product of the new zeroes is 4.

Now, we can construct the new polynomial g(x) using the sum and product of its zeroes:
g(x) = x^2 - (sum of new zeroes)x + product of new zeroes
= x^2 - 250x + 4

Thus, the polynomial whose zeroes are 2α/β and 2β/α is g(x) = x^2 - 250x + 4.

To find another polynomial whose zeroes are 1/(2α + β) and 1/(2β + α), we can follow a similar process.

The sum of the new zeroes can be calculated as follows:
sum of new zeroes = 1/(2α + β) + 1/(2β + α)
= (2β + α + 2α + β)/(2αβ + α^2 + 2β^2 + βα)
= (3α + 3β)/(3αβ + α^2 + β^2)

Similarly, using the identity (α + β)^2 - 2αβ = α^2 + β^2, we can rewrite the above expression as:
sum of new zeroes = (3(α + β))/(3αβ + (α + β)^2 - 2αβ)
= (3(α + β))/(3αβ + α^2 + β^2)
= (3(α + β))/(3αβ + 1)

Therefore, the sum of the new zeroes is (3(α + β))/(3αβ + 1).

Similarly, we can calculate the product of the new zeroes:
product of new zeroes = 1/(2α + β) * 1/(2β + α)
= 1/((2α + β)(2β + α))
= 1/(4αβ + 2α^2 + 2β^2 + αβ)
= 1/(4αβ + α^2 + β^2)

Therefore, the product of the new zeroes is 1/(4αβ + α^2 + β^2).

Now, we can construct the new polynomial h(x) using the sum and product of its zeroes:
h(x) = x^2 - (sum of new zeroes)x + product of new zeroes
= x^2 - ((3(α + β))/(3αβ + 1))x + (1/(4αβ + α^2 + β^2))

Thus, the polynomial whose zeroes are 1/(2α + β) and 1/(2β + α) is h(x) = x^2 - ((3(α + β))/(3αβ + 1))x + (1/(4αβ + α^2 + β^2)).

To find another polynomial with the given zeroes, we can use Vieta's formulas. Vieta's formulas state that for a polynomial of the form f(x) = ax^2 + bx + c, where alpha and beta are the roots, the sum of the roots is -b/a, and the product of the roots is c/a.

Let's find the sum and product of the roots alpha and beta for the polynomial f (x) = 5x^2 - 7x + 1.

The sum of the roots (alpha + beta) can be found by using the formula -b/a. In this case, a = 5 and b = -7. So, the sum of the roots is (-(-7))/5 = 7/5.

The product of the roots (alpha * beta) can be found by using the formula c/a. In this case, a = 5 and c = 1. So, the product of the roots is (1/5).

Now, let's find the polynomial whose zeroes are 2alpha/beta and 2beta/alpha.

The sum of the new zeroes (2alpha/beta + 2beta/alpha) can be found by substituting the values of alpha and beta into the formula. The sum, in this case, can be written as (2alpha * alpha + 2beta * beta)/(alpha * beta). Simplifying this expression will give us the sum of the new zeroes.

Similarly, the product can be found by substituting the values of alpha and beta into the formula and simplifying.

Once we have the sum and product of the new zeroes, we can use Vieta's formulas to find the new polynomial.

Now, let's find the polynomial whose zeroes are 1/2alpha + beta and 1/2beta + alpha.

Again, the sum of the new zeroes (1/2alpha + beta + 1/2beta + alpha) can be found by substituting the values of alpha and beta into the formula and simplifying.

Similarly, the product can be found by substituting the values of alpha and beta into the formula and simplifying.

Once we have the sum and product of the new zeroes, we can use Vieta's formulas to find the new polynomial.

In summary, to find the polynomial with the given zeroes, we need to calculate the sum and product of the old zeroes, and then use those values to find the new polynomial using Vieta's formulas.