Asked by Diya K Vineeth
If alpha and beta are the zeros f(x) = x square +5x +K
Such that alpha minus beta =1 find k
Such that alpha minus beta =1 find k
Answers
Answered by
Diya K Vineeth
f(x) =xsquare -5x +K
Here,a=1 b=-5 c=k
Alpha+beta =-b/a
=5
Apha*beta =c/a
= K
Given alpha minus beta =1
Squaring both sides
We get,. ( Alpha minus beta)^2 =1^2
Alpha^2- 2alpha*beta+beta^2=1
(Alpha^2-2alpha*beta +beta^2)-4alpha*beta = 1
(Alpha + beta)^2-4alpha*beta=1
(5)^2- 4(k). = 1
25-4k=1
-4k=1-25
K=-24/-4
K=6
Thus the value of K is 6
Here,a=1 b=-5 c=k
Alpha+beta =-b/a
=5
Apha*beta =c/a
= K
Given alpha minus beta =1
Squaring both sides
We get,. ( Alpha minus beta)^2 =1^2
Alpha^2- 2alpha*beta+beta^2=1
(Alpha^2-2alpha*beta +beta^2)-4alpha*beta = 1
(Alpha + beta)^2-4alpha*beta=1
(5)^2- 4(k). = 1
25-4k=1
-4k=1-25
K=-24/-4
K=6
Thus the value of K is 6
Answered by
Reiny
For easier typing and reading I will let alpha be "a" and beta be "b"
so you want x^2 + 5x + K = 0
-- sum of roots = a+b = -5/1 = -5
-- product of roots = k(1) = ab
given:
a - b = 1
a + b = -5
add them:
2a = -4
a = -2, then b = -3
and k = ab = (-2)(-3) = 6
you had that, good job
so you want x^2 + 5x + K = 0
-- sum of roots = a+b = -5/1 = -5
-- product of roots = k(1) = ab
given:
a - b = 1
a + b = -5
add them:
2a = -4
a = -2, then b = -3
and k = ab = (-2)(-3) = 6
you had that, good job
Answered by
oobleck
since x = -b/2a ±√(b^2-4ac)/2a
a-b = 2±√(b^2-4ac)/2a = ±√(b^2-4ac)/a = 1
b^2-4ac = a^2
25-4k = 1
k = 6
a-b = 2±√(b^2-4ac)/2a = ±√(b^2-4ac)/a = 1
b^2-4ac = a^2
25-4k = 1
k = 6
Answered by
Anonymous
How did you get _4alpha beta
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