Question

Given that α and β are the roots of the equation 2x²-3x+4=0,find an equation whose roots are α+1/α and β+1/β

Answer 1

Answer:

4x²-9x+24=0

Step-by-step explanation:

2x²-3x+4=0

ax²+bx+c=0

a=2, b=-3, c=4

sum of roots. =α+β = -b/a =3/2

product of roots. =αβ. = c/a =4/2

(α+1/α) + (β+1/β) =(α+ β) + 1/α +1/β

=(α+ β) + (β+α)/αβ

=(α+ β) + (α+β)/αβ

put the value of α+β and αβ from above

=(3/2) + (3/2)/(4/2)

=(3/2) [1 + (1)/(4/2)]

=(3/2) [1 + (2/4)]

=(3/2)[ (4+2)/4 ]

=(3/2)[ 6/4 ]

=(18/8)

=(9/4)

(α+1/α)(β+1/β) =(α)β +(α)1/β+(1/α)β +(1/α)1/β

=(αβ)+[ α/β + β/α ] +1/(αβ)

=(αβ)+[ (α^2+β^2)/(αβ) ] +1/(αβ)

=(αβ)+[(α^2+β^2 +2αβ-2αβ)/(αβ)]+1/(αβ)

=(αβ)+[(α+β)^2 -2αβ)/(αβ)]+1/(αβ)

=(αβ)+[(α+β)^2/(αβ) -2(αβ)/(αβ)]+1/(αβ)

=(αβ)+[(α+β)^2/(αβ) -2/(αβ)]+1/(αβ)

put the value of α+β and αβ from above

(α+1/α)(β+1/β) =(4/2)+[(3/2)^2/(4/2) -2/(4/2)]+1/(4/2)

=(4/2)+[(3/2)^2/(4/2) -2/(4/2)]+1/(4/2)

=(4/2)+[(9/4)/(4/2) -(4/4)]+(2/4)

=(4/2)+[(36/8) -(1)]+(1/2)

=(4/2)+[(9/2) -(1)]+(1/2)

=(4/2)+[ (9-2)/2 ]+(1/2)

=(4/2)+(7)/2+(1/2)

=(4/2)+(7+1/2)

=(4/2)+(8/2)

=(12)/2

=6

equation for required quadratic equation.

x²-Sx+P=0

x²-(9/4)x+6=0

4x²-9x+24=0

required quadratic equation.