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if 1 is one zero of the polynomial 3x^2-8x+2k+1 is seven times of the other, then find the zeros and value of k..
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Answered by
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let p(x)=3x^2-8x+2k+1----(1)
1 is zero of the polynomial p(x) then
p(1)=0
3*(1)^2 -8*1+2k+1=0
4-8+2k=0
2k=4
k=4/2=2
put k=2 in (1)
p(x)=3x^2-8x+5
given
zeroes are p=1 and q =seven times of p= 7
1 is zero of the polynomial p(x) then
p(1)=0
3*(1)^2 -8*1+2k+1=0
4-8+2k=0
2k=4
k=4/2=2
put k=2 in (1)
p(x)=3x^2-8x+5
given
zeroes are p=1 and q =seven times of p= 7
prajapatyk:
If we factorize the polynomial 3x^2-8x+5 ,then the zeroes will be 1 and 5/3
5/3 is not seven times of 1
Answered by
1
Given polynomial,
p(x)=3x^2-8x+2k+1
Let' a' and 'b' be the zeroes of the given polynomial.
As per given condition we get,
a=7b....................................1
sum of the zeroes=-(-8)/3
a+b=8/3
7b+b=8/3. (from eq1)
b=1/3....................................2
product of the zeroes =(2k+1)/3
a×b=(2k+1)/3
7b×b=(2k+1)/3
7×1/3×1/3=(2k+1)/3
7/9=(2k+1)/3
7/3=2k+1
7=6k+3
k=2/3
by substituting value of k in the given polynomial we get,
=3x^2-8x+2×2/3+1
=3x^2-8x+7/3
=9x^2-24x+7.
p(x)=0
9x^2-24x+7=0
9x^2-21x-3x+7=0
(3x-1)(3x-7)=0
x=1/3 and x=7/3
Hence value of k=2/3 and zereos of the given polynomial are 1/3 and 7/3.
p(x)=3x^2-8x+2k+1
Let' a' and 'b' be the zeroes of the given polynomial.
As per given condition we get,
a=7b....................................1
sum of the zeroes=-(-8)/3
a+b=8/3
7b+b=8/3. (from eq1)
b=1/3....................................2
product of the zeroes =(2k+1)/3
a×b=(2k+1)/3
7b×b=(2k+1)/3
7×1/3×1/3=(2k+1)/3
7/9=(2k+1)/3
7/3=2k+1
7=6k+3
k=2/3
by substituting value of k in the given polynomial we get,
=3x^2-8x+2×2/3+1
=3x^2-8x+7/3
=9x^2-24x+7.
p(x)=0
9x^2-24x+7=0
9x^2-21x-3x+7=0
(3x-1)(3x-7)=0
x=1/3 and x=7/3
Hence value of k=2/3 and zereos of the given polynomial are 1/3 and 7/3.
There is a just small mistake in it
and left everthing is alright
just i find k value , remaining i wrote the same which is quoted in the problem
I think by seeing my answer you may identify it
let us wait for manisha's response
iam sure about it because i solved many like that
but why 1 is given as zero of polynomial
That'sthe mistake occured here
then , if u want to report my answer, u can
no i don't like all those to do
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