Question

12. If a and ß are the zeros of the quadratic polynomial p(s) = 3s2 – 6s + 4, find the value of
a/B+B/a+2[1/alpha +1/beta] +3alpha×beta
​

Answer 1

Given :-

• α and ß are the zeros of the quadratic polynomial p(s) = 3s² – 6s + 4

To find:-

• Value of α/β + β/α + 2(1/α +1/β) +3αβ .

Solution:-

We know,

→α+ β = -b/a

→αβ = c/a

Given,

=>  a = 3

=> b=-6

=> c= 4  

Now,

⇒ α+ β = 6/3 = 2

⇒ αβ =4/3

Also,

⇒ α+β=2

Squaring on both sides,

⇒ α² + β² + 2αβ = 4

⇒ α²+β² = 4-2αβ =  4 - 2(4/3)

⇒α²+β² = 4/3 .

Now, we need to find the value of :

→ α/β + β/α + 2(1/α +1/β) +3αβ .

This becomes,

→α²+β² /αβ  + 2 (α+β/αβ) + 3αβ

Now, substitute the value of α²+β² in this equation.

⇒ α²+β² /αβ  + 2 (α+β/αβ) + 3αβ

⇒ 1 + 4×3 /4  +4

⇒ 1 + 3 + 4 .

⇒ 8.

Hence, value of  α/β + β/α + 2(1/α +1/β) +3αβ  is 8 .

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Answer 2

Given :-

α and ß are the zeros of the quadratic polynomial p(s) = 3s² – 6s + 4

To find:-

Value of α/β + β/α + 2(1/α +1/β) +3αβ .

Solution:-

We know,

→α+ β = -b/a

→αβ = c/a

Given,

=>  a = 3

=> b=-6

=> c= 4  

Now,

⇒ α+ β = 6/3 = 2

⇒ αβ =4/3

Also,

⇒ α+β=2

Squaring on both sides,

⇒ α² + β² + 2αβ = 4

⇒ α²+β² = 4-2αβ =  4 - 2(4/3)

⇒α²+β² = 4/3 .

Now, we need to find the value of :

→ α/β + β/α + 2(1/α +1/β) +3αβ .

This becomes,

→α²+β² /αβ  + 2 (α+β/αβ) + 3αβ

Now, substitute the value of α²+β² in this equation.

⇒ α²+β² /αβ  + 2 (α+β/αβ) + 3αβ

⇒ 1 + 4×3 /4  +4

⇒ 1 + 3 + 4 .

⇒ 8.

Hence, value of  α/β + β/α + 2(1/α +1/β) +3αβ  is 8 .

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