Question

Do THIS Find the zeroes of the quadratic polynomials given below. Find the sum and product of the zeroes and verify relationship to the coefficients of terms in the polynomial. (i) p(x) = x2 - x - 6 (ii) p(x) = x2 - 4x + 3 (iii) p(x) = x2 - 4 (iv) p(x) = x² + 2x + 1​

Answer 1

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Answer 2

Step-by-step explanation:

(i) 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