If the zeroes of the polynomial x^2-3x^2+x+1 are a-b,a,a+b, then find a and b
JessicaMalik:
It is x^3
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1
Your questions is wrong... I think it should be x^3 ... Coz three zeroes are given
Yaa ty I got it..
my ans is right?
It should be 1 instead of -1
means my ans is wrong?
Yes the value of a is wrong
ok 50 50
right
Ya
Yaa ... Alpha + beta + gamma is -b/a ... B ka value is -3 ...
ok
Answered by
1
X³-3x²+x+1
The zeroes of the polynomial are (a+b),a, (a-b)
Let α=a-b
β=a
γ=a+b
α+β+γ=-x² coefficient/x³ coefficient
a-b+a+a+b=-3/1
3a=-3
a=-3/3
a= -1
αβγ= -constant/x³ coefficient
(a-b)(a)(a+b)=-1/1
(a²-b²)(a)= -1
[(-1)²-b²](-1)=-1
(1-b²)= -1/-1
1-b²=1
1-1=b²
b²=0
b=√0=0
Therefore, a=-1 and b=0
The zeroes of the polynomial are (a+b),a, (a-b)
Let α=a-b
β=a
γ=a+b
α+β+γ=-x² coefficient/x³ coefficient
a-b+a+a+b=-3/1
3a=-3
a=-3/3
a= -1
αβγ= -constant/x³ coefficient
(a-b)(a)(a+b)=-1/1
(a²-b²)(a)= -1
[(-1)²-b²](-1)=-1
(1-b²)= -1/-1
1-b²=1
1-1=b²
b²=0
b=√0=0
Therefore, a=-1 and b=0
Its wrong ... A should be +1
Yess..but still ty!!
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