Question

If alpha and beta are zeroes of the quadratic polynomial x^2-5, then form a quadratic polynomial whose zeroes are 1+alpha and 1+beta​

Answer 1

Answer:

quadratic polynomial whose zeros are 1 + α and 1 + β will be x² - 2x - 4

Step-by-step explanation:

Given ,

x² - 5 = 0

=> (x-√5)(x + √5) = 0

=> x = ±√5

hence

α = √5 and β = -√5

new roots

= 1 + α and 1 + β

= 1 + √5 and 1 - √5

hence the new equation will be

[x-(1+α)][x-(1+β)] = 0

=> [x-(1+√5)][x-(1-√5)] = 0

=> x² -(1+√5)x - (1-√5)x + (1+√5)(1-√5) = 0

=> x² - 2x + 1 - 5 = 0

=> x² - 2x - 4 = 0

Answer 2

Answer:

If the polynomial is x2 + 5, then to calculate the zeroes you would write:

x2 + 5 = 0

which leads to

x2 = -5

which yields solutions

x = +/- i√5 where i = √-1

We are to calls these zeroes α and β. So, α = i√5 and β = -i√5.

We now want to form a polynomial with roots 1 + α and 1 + β. This yields the equation,

[x - (1 + α)][x - (1 + β)] = 0

Multiplying we get

x2 -(1 + β)x - (1 + α)x + (1 + α)(1 + β) = 0

or

x2 - (2 + α + β)x + (1 + α)(1 + β) = 0

or

x2 - (2 + i√5 - i√5) + (1 + i√5)(1 - i√5) = 0

or

x2 - 2x + 6 = 0

Step-by-step explanation:

have a nice day...