Cales larks Each.(4 x6-2(18) lf α and β are the zeroes of the polyno

1.	Cales larks Each.(4 x6-2(18) lf α and β are the zeroes of the polynomial x2 + 4x + 3, find the polynomialwhile zeroes are 1 and 1+2
Answer» Given equation = x² + 4x + 3 Factorising it by middle term splitting :-x² + 4x + 3x² + 3x + x + 3x ( x + 3 ) + 1 ( x + 3 )( x + 3 ) ( x + 1 ) ( x + 3 ) = 0x = ( - 3 ) ( x + 1 ) = 0x = ( - 1 ) So, alpha = - 3, beta = - 1 • The zeros of new equation are :- (1+alpha)/beta and (1+beta)/alpha (1+alpha)/beta = (1-3)/-1 = 2 (1+beta)/alpha = (1-1)/-3 = 0 So, the Zeros of new equation are 2 and 0 • Sum of the Zeros are :-0 + 2 = 2 • Product of the Zeros are :-0 × 2 = 0 To form the quadratic equation we have formula as :-x² - ( sum of Zeros )x + (product of Zeros) Putting value in it, we get x² - 2x + 0 So, the required quadratic equation is x² - 2x

Cales larks Each.(4 x6-2(18) lf α and β are the zeroes of the polynomial x2 + 4x + 3, find the polynomialwhile zeroes are 1 and 1+2

Answer»

Given equation = x² + 4x + 3

Factorising it by middle term splitting :-x² + 4x + 3x² + 3x + x + 3x ( x + 3 ) + 1 ( x + 3 )( x + 3 ) ( x + 1 )

( x + 3 ) = 0x = ( - 3 )

( x + 1 ) = 0x = ( - 1 )

So, alpha = - 3, beta = - 1

• The zeros of new equation are :- (1+alpha)/beta and (1+beta)/alpha

(1+alpha)/beta = (1-3)/-1 = 2

(1+beta)/alpha = (1-1)/-3 = 0

So, the Zeros of new equation are 2 and 0

• Sum of the Zeros are :-0 + 2 = 2

• Product of the Zeros are :-0 × 2 = 0

To form the quadratic equation we have formula as :-x² - ( sum of Zeros )x + (product of Zeros)

Putting value in it, we get x² - 2x + 0

So, the required quadratic equation is x² - 2x