Question

Q6. If a and B are zeroes of the polynomial 3x ^ 2 - x - 2 find a quadratic polynomial whose zeroes are 2a and 2beta​

Answer 1

Answer:

Step-by-step explanation:

Note :

Here I am using m and n instead of Alfa and beta.

Given:

If m and n are the zeroes of

quadratic polynomial f(x)=x²-3x-2

To find :

Quadratic polynomial whose

zeroes are 1/(2m+n) and 1/(2n+m) .

Explanation:

Compare f(x)= x²-3x-2 with

ax²+bx+c, we get

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

i) m+n = -b/a = -(-3)/1 = 3--(1)

ii) mn = c/a = -2/1 = -2--(2)

iii) m²+n² = (m+n)²-2mn

= 3² - 2×(-2)

= 9 + 4

= 13 ----(3)

Now ,

1/(2m+n) , 1/(2n+m) are zeroes

of a polynomial.

1) Sum of the zeroes

= 1/(2m+n) + 1/(2n+m)

= [2n+m+2m+n]/[(2m+n)(2n+m)]

= [3m+3n]/[4mn+2m²+2n²+mn]

= [3(m+n)]/[2(m²+n²)+5mn]

= [3×3]/[2×13 + 5(-2)]

= 9/(26-10)

= 9/16 ----(4)

2) Product of the zeroes

= [1/(2m+n) × 1/(2n+m)]

= 1/( 4mn+2m²+2n²+mn)

= 1/[2(m²+n²) + 5mn ]

= 1/[ 2×13 + 5(-2)]

= 1/(26-10)

= 1/16 ---(5)

______________________

Form of a quadratic polynomial

is

k[x²-(sum of the zeroes)x+product of the zeroes]

______________________

Here ,

Required polynomial is

k[ x²-(9/16)x+1/16]

For all real values of k it is true.

let k = 16,

16[x²-(9/16)x+1/16]

= 16x²-9x+1

Therefore,

Polynomial whose zeroes are

1/(2m+n) and 1/(2n+m) is

16x²-9x+1

•••••

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