Question

If $\alpha and \beta$ are the zeros of the quadratic polynomial $f(x)= x^{2} -3x-2,$,find the quadratic polynomial whose zeros are $\frac{1}{2\alpha+\beta} and \frac{1}{2\beta+\alpha}$

Answer 1

SOLUTION :

Given : α and β are the zeroes of the quadratic polynomial  f(x)= x² - 3x - 2

On comparing with ax² + bx + c,

a = 1 , b = -3  , c = -2

Sum of the zeroes = −coefficient of x / coefficient of x²

α + β  = -b/a = -(-3)/1 = 3

α + β = 3 ……………………….(1)

Product of the zeroes = constant term/ Coefficient of x²

αβ = c/a = -2/1 = -2

α×β = - 2 ……………… (2)

A.T.Q  

Sum of the zeroes of the required polynomial  = 1/2α+β + 1/ 2β+α

= (2β + α +2α + β) / (2α + β)(2β + α) [By taking L.C.M]

= ( α +2α + 2β + β) / (2α + β)(2β + α)  

= 3α + 3β /  2β (2α + β)  + α (2α + β)

= 3α + 3β /  4βα + 2β²   + 2α² + αβ

= 3α + 3β /   2β²   + 2α² + 4βα + αβ

= 3(α + β) /   2(α² + β²)  + 5βα  

= 3(α + β) / 2[(α+β)² –2αβ]  + 5βα  

[ a² + b² =  (a + b)² - 2ab  ]

=(3 × 3)  / 2[(3)² –2× -3 ]  5×(-2)]

[ From eq 1 & 2]  

= 9 / 2 [ 9 + 6 ] –10

= 9/ 2  [13 ] –10

= 9 / 26–10  

1/2α + β + 1/ 2β + α = 9/16 ……………………(3)

Product of the zeroes = 1/2α + β  × 1/ 2β + α

= 1 / (2α+β)(2β+α)

= 1/ 4αβ + 2 α² + 2β²  + αβ

= 1/ 5αβ + 2(α² + β²)

= 1 / 5αβ +2[(α+β)² –2αβ]

[ a² + b² =  (a + b)² - 2ab  ]

= 1/ 5×(−2) + 2 [(3)² –2×(−2)]

[ From eq 1 & 2]  

= 1/ −10 + 2 [ 9 +4]  

= 1/ -10 + 2[13]

= 1 / - 10 + 26

1/2α + β  × 1/ 2β + α = 1 / 16 ………………….(4)

So, the quadratic polynomial is,

kx²–(sum of the zeroes)x + (product of the zeroes)

= k (x² + 9x/ 16 + 1/16)

[ From eq 3 & 4 ]

Hence, the required quadratic polynomial is  

k(x² + 9x/ 16 + 1/16)

[K is any non zero real number]

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Answer:

x² - (9/16)x + (1/16)

Step-by-step explanation:

Given that α and β are zeroes of quadratic equation.

Given f(x) = x² - 3x - 2.

Here, a = 1, b = -3, c = -2.

(i)

Sum of zeroes = -b/a

⇒ α + β = -(-3)/1

⇒ α + β = 3.

(ii)

Product of zeroes = c/a

⇒ αβ = -2

Now,

(1) Sum of roots:

⇒ (1/2α + β) + (1/2β + α)

⇒ (2β + α + 2α + β)/(2α + β)(2β + α)

⇒ (3α + 3β)/(4αβ + 2α² + 2β² + αβ)

⇒ 3(α + β)/2(α + β)² + αβ

⇒ 3(3)/2(3)² -2

⇒ 9/16.

(2) Product of roots:

⇒ (1/2α + β)(1/2β + α)

⇒ (1/4αβ + 2α² + 2β² + αβ)

⇒ 1/2(α+β)²+αβ

⇒ 1/2(3)²-2

⇒ 1/16.

Required Quadratic Polynomial = x² - (Sum of roots)x + (Product of roots)

⇒ x² - (9/16)x + (1/16) = 0

⇒ 16x² - 9x + 1 = 0

Hope it helps!