Verify that the numbers given along side of the cubic polynomials below are their zeros. Also, verify the relationship between the zeros and coefficients in each case:
(i) 
(ii) 
Answers
SOLUTION :
(i) Given : f(x)= 2x³ + x² - 5x + 2 ;1/2,1,−2
On comparing with ax³ + bx² + cx + d ,
a = 2 , b= 1 ,c = -5 , d = 2
(a) On putting x = 1/2 in the above polynomial
f(1/2) = 2(1/2)³ + (1/2)² –5(1/2) + 2
= 2 (1/8) + ¼ – 5/2 + 2
= ¼ + ¼ - 5/2 + 2
= 2/4 - (5 + 4)/2
= ½ -½
= 0
(b) On putting x = 1 in the above polynomial,
f(1) = 2(1)³ + (1)² – 5(1) + 2
= 2 + 1 – 5 + 2
= 3 - 5 + 2
= -2 + 2
= 0
(c) On putting x = -2 in the above polynomial,
f(−2) = 2(-2)³ + (−2)² -5(−2) + 2
= 2 ×-8 + 4 +10 +2
= -16 + 4 + 10 + 2
= -16 + 14 + 2
= -16 + 16
= 0
Hence, 1/2,1,−2 are the Zeroes of the polynomial 2x³ + x² - 5x + 2 .
Let α,β,γ are the three Zeroes of the polynomial.
α = ½, β = 1, γ= -2
Sum of zeroes =−coefficient of x² / coefficient of x³
α + β + γ = −b/a
½ + 1−2 = −1/2
½ - 1 = -½
(1 - 2)/2 = -½
-½ = -1/2
Product of the zeroes = - constant term / coefficient of x²
α β γ = - d/a
½ × 1 × -2 = -2/2
½ × -2 = -1
-1 = -1
sum of the product of its zeroes taken two at a time = coefficient of x / coefficient of x³
αβ+βγ+αγ = c/a
½ × 1 + 1×(−2) +(-2)×1/2 = -5/2
½ + (-2) -1 = -5/2
½ -2 -1 = -5/2
½ -3 = -5/2
(1 - 6)/2 = -5/2
−5/2 = = -5/2
Hence, relationship between the Zeroes and the coefficients is verified.
(ii) Given : g(x) = x³ - 4x² + 5x - 2; 2,1,1
On comparing with ax³ + bx² + cx + d ,
a = 1 , b= -4 ,c = 5 , d =- 2
(a) On putting x = 2 in the given polynomial,
g(2) = (2)³ - 4(2)² + 5(2) - 2
= 8 – 16 + 10 – 2
= - 8 + 8
= 0
(b) On putting x = 1 in the given polynomial,
g(1) = (1)³ - 4(1)² + 5(1) –2
= 1 – 4 + 5 – 2
= -3 +3
= 0
Hence, 2,1,1 are the Zeroes of the polynomial x³ - 4x² + 5x - 2.
Let α,β,γ are the three Zeroes of the polynomial.
α = 2, β = 1, γ =1
Sum of zeroes = −coefficient of x² / coefficient of x³
α + β + γ = −b/a
2+1+1=−(−4)/1
4 = 4
Product of the zeroes = - constant term / coefficient of x²
α β γ = - d/a
αβγ = –(−2)
2×1×1 = 2
2 = 2
sum of the product of its zeroes taken two at a time = coefficient of x / coefficient of x³
αβ+βγ+αγ = ca
2×1+ 1×1 + 1×2 = 5/1
2 + 1 + 2 = 5
5 = 5
Hence, relationship between the Zeroes and the coefficients is verified.
HOPE THIS ANSWER WILL HELP YOU....
Answer:
We have,
f(x)=x
3
−12x
2
+39x+k
Since, roots of this equation are in A.P.
Let a−d,a,a+d are roots.
Now, sum of roots =
a
−b
a−d+a+a+d=
1
12
3a=12
a=4
Sum of products of two consecutive roots =
a
c
(a−d)a+a(a+d)+(a−d)(a+d)=
1
39
a
2
−ad+a
2
+ad+a
2
−d
2
=39
3a
2
−d
2
=39
3×16−d
2
=39
48−d
2
=39
d
2
=
d=±3
Therefore, the roots are 1, 4, 7 or 7, 4, 1
Now, product of roots=(a−d)a(a+d)=
a
−d
1×4×7=−k
k=−28
Hence, the value of k is −28.
Hence, option b is the correct option.