Question

pls
answer fast... pls..​

Answer 1

Answer:

6x²+8x+1=0

36x²-52x+1=0

x²+(-8+2√10)x-16√10=0

Step-by-step explanation:

given

sum of zeros=α+β=-8

product of sums=αβ=6

1)polynomial whose roots are 1/α and 1/β

sum of zeroes=1/α+1/β=α+β/αβ=-8/6=-4/3

product of zeroes=1/α×1/β=1/αβ=1/6

The formula of quadratic equation=x²-(α+β)x+αβ=0

x²-(-4/3)x+1/6=0

6x²+8x+1=0

2)polynomial whose roots are 1/α² and 1/β²

sum of zeroes=1/α²+ 1/β²=α²+β²/(αβ)²=(α+β)²-2αβ/(αβ)²

(-8)²-2(6)/6²=64-12/36=52/36

product of zeroes=1/α²× 1/β²=1/(αβ)²=1/6²=1/36

substitute in formula

x²-(52/36)x+1/36=0

36x²-52x+1=0

3)polynomial whose roots are (α+β),(α-β)

find α and β

α+β=-8

αβ=6

substitute in formula

x²+8x+6=0

find roots=-b±√b²-4ac/2a=-8±√64-4*1*6/2=-8±√40/2=-4±√10

α=-4+√10,β=-4-√10

sum  of roots=α+β+α-β=2α=2(-4+√10)=-8+2√10

product of roots=(α+β)×(α-β)=α²-β²=(-4+√10)²-(-4-√10)²=-16√10

substitute

x²+(-8+2√10)x-16√10=0