If the zeros of the polynomial 
are in A.P., find them.
Answers
SOLUTION :
Let, α = a - d, β = a and γ = a + d be the zeroes of the polynomial.
Given : f(x) = 2x³ - 15x² + 37x - 30
On comparing with ax³ + bx² + cx + d ,
a = 2 , b= -15 ,c = 37 , d = - 30
Sum of zeroes = −coefficient of x² / coefficient of x³
α + β + γ = −b/a
α + β + γ = -(-15/2) = 15/2
(a – d) +( a) + (a + d) = 15/2
a + a + a -d -d = 15/2
3a = 15/2
a = 15/2 × ⅓ = 5/2
a = 5/2 ………………….(1)
product of zeroes = - constant term /coefficient of x³
αβγ = -d/a
αβγ = -(-30) /2 = 30/2 = 15
αβγ = 15
(a - d) ( a) (a + d) = 15
(a - d) (a + d) (a) = 15
(a² - d²)× a = 15
on putting the value of a= 5/2 from eq 1
[(5/2)² - d² )] × 5/2 = 15
5/2 [(25/4) - d²] = 15
[(25/4) - d²] = 15 × ⅖
[(25/4) - d²] = 3×2
[(25/4) - d²] = 6
25/4 - d² = 6
25/4 - 6 = d²
(25 - 24) /4 = d²
¼ = d²
d =√1/4 = ½
d = ½ ……………….(2)
Now Substitute the value of a & d from eq 1 & 2 in α = a - d, β = a and γ = a + d to find the value of Zeroes
α = a - d = 5/2 - ½ = (5-1)/2 = 4/2 = 2
α = 2
β = a = 5/2
β = 5/2
and γ = a + d = 5/2 + ½ = (5+1)/2 = 6/2 = 3
γ = 3
Hence, the Zeroes of the polynomial are 2,5/2 & 3 .
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.