find a quadratic polynomial whose zeroes are -2/√3 , 3/√4
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Answer:
roots are -2/√3 and 3/√4
let us suppose, the rootsare α and β
therefor the equation will be
x^2 - (α + β )x + α.β = 0
Therefor α + β = -2/√3 + 3/√4
=( -2 x 2 + 3√3)/2√3
= (- 4 + 3√3)/2√3
α.β = -2/√3 x 3/√4
= -6 / 2√3
= -3 / √3
= -√3
therefor substitute the values of( α+β) and α.β in equation
x^2 - (α + β )x + α.β = 0
x^2 - ((- 4 + 3√3)/2√3)x + (-√3)
x^2 + (( 4 - 3√3 )/2√3)x - (√3 )
The equation is
x^2 + (( 4 - 3√3 )/2√3)x - (√3 )
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