Question

1. Find the zeors of x²+2x+3 and hence
find their sum and product.
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Answer 1

Answer:

x² - x - 6

= x² - 3x + 2x - 6

= x(x - 3) +2(x -3)

= (x - 3)(x +2)

now,P(x) = 0, then, x = -2 and 3

hence,zeros of P(x) is -2 and 3

sum of zeros = - coefficient of x/coefficient of x² =

LHS = (-2 + 3) = 1

RHS = -(-1)/1 = 1

So, LHS = RHS

similarly, product of root = constant/coefficient of x²

LHS =(-2) × 3 =-6

RHS = -6/1 = -6

so, LHS = RHS

hence, verified

(ii) x² - 4x + 3

= x² - 3x - x + 3

= x(x - 3) - 1(x - 3)

= (x - 1)(x - 3)

now, P(x) = 0, then, x = 1 and 3

hence, zeros of P(x) are 1 and 3

now sum of zeros = - coefficient of x/coefficient of x²

LHS = 1 + 3= 4

RHS = -(-4)/1 = 4

so, LHS = RHS

similarly, product of zeros = constant/coefficient of x²

LHS = 1 × 3 =3

RHS= 3/1 = 3

so, LHS = RHS

hence, verified

(iii) x² - 4

= (x - 2)(x + 2)

now, P(x) = 0, then , x=-2 and 2

sum of zeros = - coefficient of x/coefficient of x²

LHS =-2 + 2 = 0

RHS= 0/1 = 0

so, LHS = RHS

similarly, product of zeros = constant/coefficient of x²

LHS = -2 × 2 =-4

RHS = -4/1 = -4

so,LHS = RHS

hence, verified

(iv) x² + 2x + 1

= (x + 1)(x + 1)

now, P(x) =0 , then,x = -1 and -1

now, sum of zeros = - coefficient of x/coefficient of x²

LHS = -1 - 1 = -2

RHS =-(2)/1 = -2

so, LHS = RHS

similarly product of zeros = constant/coefficient of x²

LHS = -1 × -1 = 1

RHS = 1/1 = 1

so, LHS = RHS

hence, verified

Step-by-step explanation:

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Answer 2

Answer:

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