if alpha and beta are the zeros of the polynomial f x is equal to x square - 5 x + K such that Alpha minus beta equal to 1 find the value of k
123Varun11:
k = 6
Answers
Answered by
107
here is your answer by Sujeet,
given that,
x²-5x+k
sum of zeroes =-b/a
a+b=-(5)/1
a+b=5
product of zeroes=c/a=k/1
Now solving the equation,
a+b=5
a-b=1
_____
2b=4
b=4/2
b=2
Now,
a-b=1
a-2=1
a=3
Again,
ab=c/a
3*2=k/1
k=6
that's all
Sujeet
mark brainliest
given that,
x²-5x+k
sum of zeroes =-b/a
a+b=-(5)/1
a+b=5
product of zeroes=c/a=k/1
Now solving the equation,
a+b=5
a-b=1
_____
2b=4
b=4/2
b=2
Now,
a-b=1
a-2=1
a=3
Again,
ab=c/a
3*2=k/1
k=6
that's all
Sujeet
mark brainliest
Answered by
75
Given f(x)=x²-5x+k
And α,β are the roots of f(x).
Also given that the difference of the roots i.e.,α-β=1-----(1)
Discriminant(Δ) of f(x)=b²-4ac=(-5)²-4(1)(k)=25-4k
For a quadratic equation ax²+bx+c=0,Difference of roots=(√Δ)/a
⇒α-β=[√(25-4k)]/1
But from (1),α-β=1
⇒1=√(25-4k)
⇒25-4k=1
⇒4k=24
⇒k=6.
And α,β are the roots of f(x).
Also given that the difference of the roots i.e.,α-β=1-----(1)
Discriminant(Δ) of f(x)=b²-4ac=(-5)²-4(1)(k)=25-4k
For a quadratic equation ax²+bx+c=0,Difference of roots=(√Δ)/a
⇒α-β=[√(25-4k)]/1
But from (1),α-β=1
⇒1=√(25-4k)
⇒25-4k=1
⇒4k=24
⇒k=6.
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