Question

Find the zeros of the quadratic polynomials given below. Find the sum and product of the zeros 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

(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

Answer 2

Dear Student.

Solution :

We know that standard quadratic polynomial ax² + bx +c

let the zeros are α and β .

So, α + β = -b/a

αβ = c/a

(i) p(x) = x² – x – 6

x² – 3x + 2x – 6 =0

x(x-3) + 2 ( x-3) =0

(x- 3)( x+2) =0

x = 3 and x =-2

From coefficients α + β = -b/a = 1 ------ eq 1

from obtained zeros 3-2 = 1 --------- eq 2

sum of zeros are equal ;eq 1 and eq2 are equal.

αβ = c/a = -6

3(-2) = -6

(ii) p(x) = x² – 4x + 3

x² – 3x - x + 3 =0

x( x-3) -1 ( x- 3) =0

(x - 3)( x -1) =0

x = 3 and x = 1

α + β = -b/a = 4

3+ 1 = 4

αβ = c/a = 3

3(1) = 3

(iii) p(x) = x² – 4

x² – 4 =0

(x+2)( x- 2) =0

x = -2 and x = 2

α + β = -b/a = 0

-2 + 2 = 0

αβ = c/a = -4

2(-2) = -4

(iv) p(x) = x² + 2x + 1

x² + x + x + 1 =0

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

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

x = -1 and x = -1

α + β = -b/a = -2

-1 -1 = -2

αβ = c/a = 1

(-1)(-1) = 1

Hope it helps you.