Question

find a quadratic polynomial whose zeroes are 1/2+2 root3 and 1/2_2 root3

Answer 1

hi
here α=1/2+2√3 β=1/2-2√3
quadratic polynomial is
x^2+(α+β)x-αβ=0
so
x^2+(1/2+2√3+1/2-2√3)x-(1/2+2√3)(1/2-2√3)=0

Answer 2

Hiii friend,

Let Alpha = 1/2+2✓3 and beta = 1/2-2✓3


Therefore,

Sum of zeros = (Alpha + Beta) = (1/2+2✓3 + 1/2-2✓3 = 2+2✓3+2-2✓3/(2+2✓3)(2-2✓3) = 4/(2)² - (2✓3)² = 4/ 4 - 12 = 4/-8


Product of zeros = (Alpha × Beta) = 1/2+2✓3 × 1/(2-2✓3) = 1/(2+2✓3)(2-2✓3) =1/(2)² - (2✓3)² = 1/4 - 12 = 1/-8



Therefore,


Required polynomial = X²-(Alpha + Beta)X + Alpha × Beta


=> X²-(4/-8)X + 1/(-8)

=> X²-4X/-8 +1/-8


=> -8X²-4X+1 = 0



HOPE IT WILL HELP YOU...... :-)