If the zeroes of the polynomial f(x) = x³ –12x² + 39x + k are in A.P., find the value of k.
Answers
SOLUTION :
Let, a-d, a, a+d be the zeroes of the polynomial f(x).
Given : f(x) = x³ –12x² + 39x + k
On comparing with ax³ + bx² + cx + d ,
a = 1 , b = -12 ,c = 39 , d = k
Sum of zeroes = −coefficient of x² / coefficient of x³
α + β + γ = −b/a
α + β + γ = - (-12)/1 = 12
(a – d) +( a) + (a + d) = 12
a + a + a -d -d = 12
3a = 12
a = 12/3 = 4
a = 4
Since, a is the zero of the polynomial f(x),
Therefore, f(a) = 0
f(a)= a³ –12a² + 39a + k
f(4) = 4³ –12(4)² + 39(4) + k = 0
64 – 12 × 16 + 156 + k = 0
64 - 192 +156 + k = 0
64 +156 -192 + k = 0
220 -192 + k = 0
28 + k = 0
k = -28
Hence, the value of k is -28 .
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.