P(s)=3s2−6s+4, Find the Value of α/β + β/α + 2(1/α + 1/β) + 3αβ.
The question states "If α and β are the zeros of the quadratic polynomial p(s)=3s2−6s+4, find the value of α/β + β/α + 2(1/α+1/β) + 3αβ.” At most, a polynomial function can have real roots equal to its degree. Set a function's value to zero and solve to find its roots. The steps to find the Value of α/β + β/α + 2(1/α + 1/β) + 3αβ. are as follows:
Step 1: As it is given that α and β are the zeros of the quadratic polynomial 3s2−6s+4
Sum of zeroes = α + β = -(-6)/3 = 2
Product of zeros = αβ = 4/3
Now, α/β + β/α + 2(1/α + 1/β) + 3αβ
α2+ β2/αβ + 2(1/α + 1/β) + 3αβ
(α + β)2 - 2αβ/ αβ + 2 (β + α)/αβ + 3αβ
Step 2: Substituting the values of α + β and αβ
(α + β)2 - 2αβ/ αβ + 2 (β + α)/αβ + 3αβ
2*2 - 2* (4/3)/(4/3) + 2*2/(4/3) + 3*(4/3)
1 + 12/4 + 12/3
1 + 3 + 4 = 8
Hence, the value of (α + β)2 - 2αβ/ αβ + 2 (β + α)/αβ + 3αβ = 8
Summary:
If α and β are the zeros of the quadratic polynomial p(s)=3s2−6s+4, find the value of α/β + β/α + 2(1/α+1/β) + 3αβ.
The value of α/β + β/α + 2(1/α+1/β) + 3αβ is 8 if α and β are the zeros of the quadratic polynomial p(s)=3s2−6s+4. Polynomial operations include division, subtraction, multiplication, and addition. With the use of fundamental algebraic concepts and factorization techniques, any polynomial may be solved with ease. The first step in solving the polynomial equation is to set the right-hand side's value to zero.
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