Question

find out the polynomial whose zeroes are (3α + 1) and (3β + 1)​

Answer 1

Answer:

Given :  alpha and beta are the zeros of the polynomial x2-4x+3

To find : the quadratic polynomial whose zeroes are  1/(3α+β),  1/(3β+α)

Solution:

α &  β  are  zeros of the polynomial x²-4x+3

=> α +  β  = -(-4)/1 = 4

   α β   = 3/1  = 3

quadratic polynomial whose zeroes are 1/(3α+β),  1/(3β+α)

Sum of Zeroes =  1/(3α+β) + 1/(3β+α)

= ( 3β+α + 3α+β) / (9αβ + αβ + 3α² + 3β²)

= ( 4α+4β)/( 3(α +  β)² + 4αβ)

=  (4 ( 4) )/ ( 3 (4)² + 4(3))

= 16 / 60

Product of roots   =   1/(3α+β) *  1/(3β+α)  =   1/ (9αβ + αβ + 3α² + 3β²)

= 1/60

x² - (16/60)x + 1/60 = 0

=> 60x²  - 16x  + 1  = 0

Step-by-step explanation:

Answer 2

★ Gɪᴠᴇɴ :

• α and β are the zeroes
• x²-x-2

☢ Tᴏ Fɪɴᴅ :

• polynomial whose zeroes of (3α+1) & (3β+1)

☯ Sᴏʟᴜᴛɪᴏɴ :

↠ x² - x - 2 = 0

↠ x² - (2 - 1)x - 2 = 0

↠ x² - 2x + x - 2 = 0

↠ x(x - 2) + 1(x - 2) = 0

↠ (x - 2) (x + 1)

⠀⠀◉ (2) and (-1) are the zeroes

⠀⠀◉ Hence, α = 2 ; β = (-1)

⚘ Nᴏᴡ, Fɪɴᴅɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ ɢɪᴠᴇɴ ᴢᴇʀᴏᴇs :

→ 3α + 1 = 3 × 2 + 1 (α = 2)

→ 3α + 1 = 6 + 1 = 7

❊ Nᴏᴡ, ᴛʜᴇ sᴇᴄᴏɴᴅ ᴢᴇʀᴏ :

↬ 3β + 1 = 3 × (-1) + 1 [β = (-1)]

↬ 3β = -3 + 1 = (-2)

⠀⠀◉ α = 7 ; β = (-2)

❥ Hᴇɴᴄᴇ, ᴛʜᴇ ʀᴇϙᴜɪʀᴇᴅ ᴘᴏʟʏɴᴏᴍɪᴀʟ :

↦ f(x) = k(x²-[7+(-2)]x+[(7 × (-2)])

↦ k(x² - (7 - 2)x - 14)

↦ k(x² - 5x - 14)

☛ f(x) = k(x² - 5x - 14)