SOLVE IT PLZ CORRECT ANSWER WILL BE MARKED AS BRAINLIST ANSWER
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Let a,b,c be the zeros of the polynomial.
Given f(x) = x^3 - 5x^2 - 2x + 24.
We know that sum of roots = -b/a
a + b + c = -(-5)/1
= -(-5)
= 5. ------ (1)
We know that product of roots = -d/a
abc = -24/1
= -24. ----- (2)
Given that product of its two zeroes = 12.
ab = 12 ----- (3)
Substitute (3) in (2), we get
12 * c = -24
c = -2
Substitute c = -2 in (1),
a + b - 2 = 5
a + b = 7 ----- (4)
We know that (a + b)^2 = (a - b)^2 + 4ab
(7)^2 = (a - b)^2 + 4(12)
49 = (a - b)^2 + 48
a - b = 1. ---- (5)
on solving 4 and 5, we get
a + b = 7
a - b = 1
--------------
2b = 8
b = 4
substitute b 4 in (5), we get
a -b = 1
a = 3.
Therefore zeroes = 3,4,-2.
Given f(x) = x^3 - 5x^2 - 2x + 24.
We know that sum of roots = -b/a
a + b + c = -(-5)/1
= -(-5)
= 5. ------ (1)
We know that product of roots = -d/a
abc = -24/1
= -24. ----- (2)
Given that product of its two zeroes = 12.
ab = 12 ----- (3)
Substitute (3) in (2), we get
12 * c = -24
c = -2
Substitute c = -2 in (1),
a + b - 2 = 5
a + b = 7 ----- (4)
We know that (a + b)^2 = (a - b)^2 + 4ab
(7)^2 = (a - b)^2 + 4(12)
49 = (a - b)^2 + 48
a - b = 1. ---- (5)
on solving 4 and 5, we get
a + b = 7
a - b = 1
--------------
2b = 8
b = 4
substitute b 4 in (5), we get
a -b = 1
a = 3.
Therefore zeroes = 3,4,-2.
siddhartharao77:
:-)
thnx
why we have taken (a+b)^2?
oh(a-b)^2
its a formula
but it is not written in the question
is it necessary to find a-b?
in this question it is mandatory to find a-b..
ok
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Hi,
Please see the attached file!
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Please see the attached file!
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