Question

If $\alpha and \beta$ are the zeros of the quadratic polynomial such that $\alpha +\beta=24 and \alpha-\beta =8$, find a quadratic polynomial have $\alpha and \beta$ as its zeros

Answer 1

SOLUTION :

Given :  α and β are the zeroes of the quadratic polynomial

α + β = 24 ………… (1)  

α–β = 8 ………………..(2)

On Adding eq 1&2,  

α + β = 24

α–β = 8

-------------------

2α = 32

α= 32 / 2 = 16

α = 16

Put the value of α, in eq 2 ,  

α–β = 8

16 - β = 8

β = 16–8

β = 8

Now,

Sum of the zeroes of the required polynomial = α+β = 16 + 8 = 24

α + β  = 24 ……………..(3)

Product of the zeroes of the required polynomial = αβ = 16×8 = 128

αβ = 128 …………….(4)

Then, the quadratic polynomial is :  

Kx² –(sum of the zeroes)x + (product of the zeroes)

=k( x² - 24x + 128)

[ From eq 3 & 4 ]

[K is any non zero real number]

Hence, the required quadratic polynomial is f(x) =k (x² - 24x + 128)

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Answer 2

Answer:

SOLUTION :

Given :  α and β are the zeroes of the quadratic polynomial

α + β = 24 ………… (1)  

α–β = 8 ………………..(2)

On Adding eq 1&2,  

α + β = 24

α–β = 8

-------------------

2α = 32

α= 32 / 2 = 16

α = 16

Put the value of α, in eq 2 ,  

α–β = 8

16 - β = 8

β = 16–8

β = 8

Now,

Sum of the zeroes of the required polynomial = α+β = 16 + 8 = 24

α + β  = 24 ……………..(3)

Product of the zeroes of the required polynomial = αβ = 16×8 = 128

αβ = 128 …………….(4)

Then, the quadratic polynomial is :  

Kx² –(sum of the zeroes)x + (product of the zeroes)

=k( x² - 24x + 128)

[ From eq 3 & 4 ]

[K is any non zero real number]

Hence, the required quadratic polynomial is f(x) =k (x² - 24x + 128)

HOPE THIS ANSWER WILL HELP YOU...