Question

10th Maths:-
If $m, n$ are the zeroes of polynomial $dz^2 - ez + f$, then the polynomial $( \dfrac{d + fy}{f} )^2 - ( \dfrac{e}{f} )^2 y$ has its zeroes as:
(1) $m^{-1} , n^{-1}$
(2) $m^{-2} , n^{-2}$
(3) $m n^{-1} , m^{-1} n$
(4) $m^2 , n^2$
​

Answer 1

Option (2)

Step-by-step explanation:

Given :-

m and n are the zeroes of the polynomial dz²-ez+f .

To find :-

Find the zeroes of [(d+fy)/f]²-(e/f)²y?

Solution :-

Given that

The quardratic polynomial is dz²-ez+f

On comparing it with the standards quardratic polynomial ax²+bx+c

a = d

b = -e

c = f

Given zeroes are m and n

We know that

Sum of the zeroes = -b/a

=> m+n = -(-e)/d

=> m+n = e/d -------(1)

and

Product of the zeroes = c/a

=> mn = f/d ---------(2)

Given another quardratic polynomial is

[(d+fy)/f]²-(e/f)²y

=> [(d²+f²y²+2dfy)/f²] - (e²y/f²)

=> [(d²+f²y²+2dfy) - (e²y)]/f²

=> (d²+f²y²+2dfy-e²y)/f²

=> [f²y²+(2df-e²)y+d²]/f²

On comparing it with the standards quardratic polynomial ax²+bx+c

a = f²/f² = 1

b = (2df-e²)/f²

c = d²/f²

We know that

Sum of the zeroes = -b/a

=> -(2df-e²)/f²×1

=> (e²-2df)/f² -----------(3)

Product of zeroes = c/a

=> (d²/f²)/1

=> d²/f²

=> (d/f)²

=> (1/mn)² from (2)

=> (mn)^-2

=> m^-2 n^2 ------------(4)

From (4) we conclude that m^-2 and n^-2 are the zeroes.

Now

Sum of the zeroes= m^-2 + n^-2

=> (1/m²)+(1/n²)

=> (m²+n²)/(m²n²)

=> [(m+n)²-2mn]/(mn)²

=> [(e/d)²-2(f/d)]/(f/d)²

=> [(e²/d²)-(2f/d)]/(f²/d²)

=> [(e²-2fd)/d²]/(f²/d²)

=> (e²-2fd)/f²----------(5)

(3)&(5) are same.

Therefore, m^-2 and n^-2 are the zeroes .

Answer :-

The zeroes of [(d+fy)/f]²-(e/f)²y are m^-2 and n^-2

Used formulae:-

→ The standards quardratic polynomial is ax²+bx+c

→ Sum of the zeroes=-b/a

→ Product of zeroes = c/a