If α and β are the roots of the equation 3x2+8x+2=0 then 1α+1β=

Given quadratic equation is,3x2+8x+2=0. Also, α and β are the roots of the equation.We know that,The standard form of a quadratic equation is ax2+bx+c=0, where a,b c are real numbers and a≠0.And, the sum of the roots is given by –ba and product of the roots is given by ca.After comparing 3x2+8x+2=0 with the standard form of the quadratic equation, we get,a=3,b=8, c=2. Then, Sum of the roots =–ba=-83.Product of the roots =ca=23.Now, we have to find 1α+1β=α+βαβ -1) ⟹α+β is nothing but the sum of the roots of the equation.∴ α+β= ba=-83 -2) ⟹αβ is nothing but the product of the roots of the equation∴αβ=ca=23 -3) After substituting equations 2) & 3) in 1) we get,1α+1β = -83×32⇒1α+1β= -4 ∴ If α and β are the roots of the equation 3x2+8x+2=0 then 1α+1β=-4.Hence, the correct option is 3.

If α and β are the roots of the equation 3x2+8x+2=0 then 1α+1β=

Given quadratic equation is,3x2+8x+2=0. Also, α and β are the roots of the equation.We know that,The standard form of a quadratic equation is ax2+bx+c=0, where a,b c are real numbers and a≠0.And, the sum of the roots is given by –ba and product of the roots is given by ca.After comparing 3x2+8x+2=0 with the standard form of the quadratic equation, we get,a=3,b=8, c=2. Then, Sum of the roots =–ba=-83.Product of the roots =ca=23.Now, we have to find 1α+1β=α+βαβ -1) ⟹α+β is nothing but the sum of the roots of the equation.∴ α+β= ba=-83 -2) ⟹αβ is nothing but the product of the roots of the equation∴αβ=ca=23 -3) After substituting equations 2) & 3) in 1) we get,1α+1β = -83×32⇒1α+1β= -4 ∴ If α and β are the roots of the equation 3x2+8x+2=0 then 1α+1β=-4.Hence, the correct option is 3.