if the zeros of the polynomials x³ - 3x² +x+1 are a-b,a,a+b, find a and b
Answers
Answered by
27
hi friend ,
given the zeros of the polynomials x³ - 3x² +x+1 are a-b,a,a+b
the sum of the zeros are a-b+a+a+b=3a=-x²-coeffiecient /x³co efficient =3
→a=1
→now the zeros are 1-b,1,1+b
product of zeros = -constant term /x³ coefficient
→1-b²=-1
→b²=2
b=±√2
I hope this will help u ;)
given the zeros of the polynomials x³ - 3x² +x+1 are a-b,a,a+b
the sum of the zeros are a-b+a+a+b=3a=-x²-coeffiecient /x³co efficient =3
→a=1
→now the zeros are 1-b,1,1+b
product of zeros = -constant term /x³ coefficient
→1-b²=-1
→b²=2
b=±√2
I hope this will help u ;)
mysticd:
it is wrong. plz check.
Answered by
15
By trail and error method, let x=1
f(1) = 1-3+1+1 = 0
So, 1 is one of the zero of given polynomial
So x-1 is one of the factor of given polynomial.
So dividing the given polynomial with x-1, we will get another factor
x-1) x³-3x²+x+1 (x²-2x-1
x³- x²
(-) (+)
__________
0 - 2x²+x+1
- 2x²+2x
(+) (-)
__________
-x+1
-x+1
(+) (-)
__________
0
___________
So, x²-2x-1 is also a factor of given polynomial
Solving the factor x²-2x-1 =0.
Here, a=1, b=-2, c=-1
Roots of the equation be solved by the formula : (-b+-√b²-4ac) / 2a
= 2+-√4+4/2
= (2+-2√2)/2
= 2(1+-√2)/2
Therefore, 1+√2 and 1-√2 are the other zeroes of given polynomial.
therefore the zeroes are 1, 1+√2, 1-√2
let a=1, a-b=1-√2, a+b=1+√2
placing the value of a in a+b
a+b = 1+√2
1+b = 1+√2
b=1+√2-1
b=√2
Hence, a=1, b=√2
f(1) = 1-3+1+1 = 0
So, 1 is one of the zero of given polynomial
So x-1 is one of the factor of given polynomial.
So dividing the given polynomial with x-1, we will get another factor
x-1) x³-3x²+x+1 (x²-2x-1
x³- x²
(-) (+)
__________
0 - 2x²+x+1
- 2x²+2x
(+) (-)
__________
-x+1
-x+1
(+) (-)
__________
0
___________
So, x²-2x-1 is also a factor of given polynomial
Solving the factor x²-2x-1 =0.
Here, a=1, b=-2, c=-1
Roots of the equation be solved by the formula : (-b+-√b²-4ac) / 2a
= 2+-√4+4/2
= (2+-2√2)/2
= 2(1+-√2)/2
Therefore, 1+√2 and 1-√2 are the other zeroes of given polynomial.
therefore the zeroes are 1, 1+√2, 1-√2
let a=1, a-b=1-√2, a+b=1+√2
placing the value of a in a+b
a+b = 1+√2
1+b = 1+√2
b=1+√2-1
b=√2
Hence, a=1, b=√2
Similar questions