Find the zeroes of the quadratic polynomials given below. Find the sum and product
ofthe zeroes and verify relationship to the coefficients ofterms in the polynomial
(i) pbx) = x-x-6 (ii) p(x) = r - 4x +3
(iii) p(x)=x2-4
(iv) p(x) = x2 + 2x + 1
Answers
Step-by-step explanation:
(i) p(x) = x²-x-6
x²-3x+2x-6
x(x-3)+2(x-3)
(x-3)(x+2)
x = 3...... x = -2
sum of the zeroes = 3+(-2) = 3-2 = 1
product of zeroes = (3)(-2) = -6
(ii) p(x) = x²-4x+3
x²-3x-x+3
x(x-3)-1(x-3)
(x-3)(x-1)
x = 3 ....... x = 1
sum of the zeroes = 3+1 = 4
product of zeroes = (3)(1) = 3
(iii) p(x) = x²-4
x²-4=0
x² = 4
x = √4 = 2
sum of the zeroes = 2+2 = 4
product of zeroes = (2)(2) = 4
(iv) p(x) = x²+2x+1
x²+x+x+1
x(x+1)+1(x+1)
(x+1)(x+1)
x= -1.....x=-1
sum of the zeroes = -1+(-1) = -1-1 = -2
product of zeroes = (-1)(-1) = 1
THE RELATIONSHIP BETWEEN THE COEFFICIENTS OF TERMS AND THE ZEROES OF THE POLYNOMIAL IS
IF ALPHA AND BETA ARE THE ZEROES OF THE POLYNOMIAL THEN
ALPHA + BETA = COEFFICIENT OF X²/ COEFFICIENT OF X
ALPHA × BETA = CONSTANT TERM/ COEFFICIENT OF X
hope this helps you.....
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