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If $α$ and $β$ are zeroes of the polynomial $4 x 2 − x − 4,$ find the quadratic polynomial whose zeroes are $1 2 α$ and $1 2 β$ .
5)

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\frac{1}{8} (8 x^{2} + x - 2)

\frac{1}{4} (4 x^{2} + 4 x - 2)

2 (4 x^{2} + x - 4)

- 2 (4 x^{2} + x - 4)

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Detailed Solution

Given that

α

and

β

are zeroes of the polynomial

4 x 2 − x − 4.

We need to find the quadratic polynomial whose zeroes are

1 2 α

and

1 2 β

.
Based on the relationship between zeroes and polynomials we have quadratic polynomial as:

p (x) = x 2 − (α + β) x + α β = 0 ..... (i)

where

α & β

are the zeroes of the polynomial.
We have the formula that,
The sum of zeroes

= − Coefficient of x Coefficient of x 2

and
product of zeroes

= Constant term Coefficient of x 2

Let us assume

p x = 4 x 2 − x − 4

and,

α

and

β

be the zeroes of the polynomial

p (x)

, then

α + β = − Coefficient of x Coefficient of x 2 ⇒ α + β = − (− 1) 4 ∴ α + β = 14

And

α β = Constant term Coefficient of x 2 ⇒ α β = − 4 4 ∴ α β = − 1

Now for

1 2 α

and

1 2 β

we have,
Sum of zeroes:

1 2 α + 1 2 β = β + α 2 α β ..... (i i)

Placing the value of

β + α

and

α β

in equation

i i

1 2 α + 1 2 β = 14 2 (− 1) ⇒ 1 2 α + 1 2 β = − 1 8

And product of zeroes is:

1 2 α ⋅ 1 2 β = 1 4 α β ..... (i i i)

Placing the value of

α β

in equation

(i i i)

1 2 α ⋅ 1 2 β = 1 4 (− 1) ⇒ 1 2 α ⋅ 1 2 β = 1 − 4

Now placing the value of sum of zeroes and product of zeroes in equation

(i)

we have,

f (x) = x 2 + 18 x − 14 ⇒ f (x) = 18 (8 x + 2 x − 2)

Hence, The required quadratic polynomial is

18 (8 x + 2 x − 2)

.
Therefore option 1 is correct.

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