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Rajesh donated some money and books to a school for poor children. Money and books can be represented by the zeroes (i.e., $α, β$ ) of the polynomial $p (x) = 2 x 2 − 5 x + 7.$ Akshita who is a friend of Rajesh, also got inspired by him and donated the money and books in the form of a polynomial whose zeroes are $2 α + 3 β$ and $3 α + 2 β$ . Find the polynomial whose zeroes are $2 α + 3 β$ and $3 α + 2 β$ .

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5 x 2 − 16 x − 35 = 0

2 x 2 − 25 x + 82 = 0

− 2 x 2 + 25 x − 82 = 0

− 5 x 2 + 16 x + 35 = 0

answer is D.

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Detailed Solution

Given that the money

α

and books

β

donated by Rajesh are the zeroes of the polynomial

p (x) = 2 x 2 − 5 x + 7.

Money

2 α + 3 β

and books

3 α + 2 β

donated by Akshita are the zeroes of a polynomial.
We need to find the polynomial whose zeroes are

2 α + 3 β

and

3 α + 2 β

.
When

α

and

β

are the zeroes of quadratic polynomial, the two equations formed with zeroes are given as:

α + β = − b a

and

α β = c a

where

b

is the coefficient of linear term,

c

is constant while

a

is the coefficient of quadratic term in quadratic polynomial.
Based on the relationship between zeroes and polynomials we have quadratic polynomial as:

f (x) = x 2 − (α + β) x + α β

where

α & β

are the zeroes of the polynomial.
Now, from the polynomial

p (x) = 2 x 2 − 5 x + 7

we have,

a + β = − (− 5) 2 = 52 a β = 72

We know that Akshita's donations are

(2 α + 3 β)

and

(3 α + 2 β)

which are roots of a polynomial.
Let the polynomial be

f (x) = x 2 − 2 α + 3 β + (3 α + 2 β) x + 2 α + 3 β (3 α + 2 β) ..... (i)

Sum of zeroes is:

(2 α + 3 β) + (3 α + 2 β) = 2 α + 3 α + 3 β + 2 β ⇒ (2 α + 3 β) + (3 α + 2 β) = 5 α + 5 β ⇒ (2 α + 3 β) + (3 α + 2 β) = 5 α + β ... (i i)

Now equating the value of

a + β

in equation

(i i)

we have,

(2 α + 3 β) + (3 α + 2 β) = 552 ∴ (2 α + 3 β) + (3 α + 2 β) = 252 ... (i i i)

Similarly, product of zeroes is:

(2 α + 3 β) (3 α + 2 β) = 6 α 2 + 13 α β + 6 β 2

⇒ (2 α + 3 β) (3 α + 2 β) = 6 α 2 + 12 α β + α β + 6 β 2 ⇒ (2 α + 3 β) (3 α + 2 β) = 6 α 2 + β 2 + 2 α β + α β ⇒ (2 α + 3 β) (3 α + 2 β) = 6 (α + β) 2 + α β ... (i v)

Now equate the value of

(α + β) & (α β)

in equation

(i v)

, we have,

(2 α + 3 β) (3 α + 2 β) = 65222 + 72 ⇒ (2 α + 3 β) (3 α + 2 β) = 6 × 254 + 72 ⇒ (2 α + 3 β) (3 α + 2 β) = 752 + 72 ∴ (2 α + 3 β) (3 α + 2 β) = 822 ... (v)

Now equating the values from equation

(i i i) & (v)

into equation

(i)

f (x) = x 2 − 252 x + 822 ⇒ f (x) = 2 x 2 − 25 x + 82 2

Equating

f (x)

to zero we have,

2 x 2 − 25 x + 82 = 0

Hence, the polynomial whose zeroes are

2 α + 3 β

and

3 α + 2 β

2 x 2 − 25 x + 82 = 0

.
Therefore, option 4 is correct.

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