If the zeros of the polynomial f(x)=xcube-3xsquare+x+1 are a-b a, a+b find a and b
Answers
Question
If the zeros of the polynomial f(x) = (x³ - 3x² + x + 1 )are a-b , a, a+b find a and b
Solution
Given:-
- polynomial, f(x) = (x³-3x²+x+1) =0
- (a-b) , a and (a+b) three zeroes of this equation .
Find:-
- zeroes of this equation
- value of a and b
Explanation:-
☛ Sum of zeroes = [ -(coefficient of x²)/(coefficient of x³)]
➛ (a-b)+(a)+(a+b) = -(-3)/1
➛ 3a = 3
➛ a = 3/3
➛ a = 1
☛product of zeroes = [-(constant part)/(coefficient of x³)]
➛ (a-b)*(a)*(a+b) = -1/1
➛ (a-b)*(a)*(a+b) = -1
keep value of a = 1 in above equation,
➛ (1-b)*(1)*(1+b) = -1
➛ (1²-b²) = -1. [ (x-1)(x+1)=(x²-1²) ]
➛ b² = 1 + 1
➛ b = √2
Thus:-
- Value of a = 1
- Value of b = √2
And,
- First zero of polynomial (a-b) = (1 - √2)
- Second zero a = 1
- Third zeroes = (a+b) = (1 + √2).
☛ Zeroes of given polynomial will be (1-√2), 1 and (1 +√2) [ Ans.]
_____________________
Verification
Given equation
- x³ - 3x² + x + 1 = 0
➛ x³ - x² - 2x² + 2x - x + 1 = 0
➛ x²( x -1) - 2x ( x - 1) - 1( x - 1) = 0
➛ ( x - 1) ( x² - 2x - 1) = 0
➛ ( x - 1).(x²-x-x√2-x+1+√2+x√2-√2-2) =0
➛ (x-1).[x(x-1-√2)-1(x-1-√2)+√2(x-1-√2)] = 0
➛ (x-1).[(x-1+√2)(x-1-√2)] = 0
➛ (x-1) = 0 Or, (x-1+√2) = 0 Or, (x-1-√2) = 0
➛ x = 1, Or. x = (1-√2) Or , x = (1+√2)
Hence, zeroes of this equation are 1 , (1+√2) and (1-√2)
these are same .
So, we can say that our solution is absolutely right .
__________________
Question
If the zeros of the polynomial f(x) = (x³ - 3x² + x + 1 )are a-b , a, a+b find a and b
Solution
Given:-
polynomial, f(x) = (x³-3x²+x+1) =0
(a-b) , a and (a+b) three zeroes of this equation .
Find:-
zeroes of this equation
value of a and b
Explanation:-
☛ Sum of zeroes = [ -(coefficient of x²)/(coefficient of x³)]
➛ (a-b)+(a)+(a+b) = -(-3)/1
➛ 3a = 3
➛ a = 3/3
➛ a = 1
☛product of zeroes = [-(constant part)/(coefficient of x³)]
➛ (a-b)*(a)*(a+b) = -1/1
➛ (a-b)*(a)*(a+b) = -1
keep value of a = 1 in above equation,
➛ (1-b)*(1)*(1+b) = -1
➛ (1²-b²) = -1. [ (x-1)(x+1)=(x²-1²) ]
➛ b² = 1 + 1
➛ b = √2
Thus:-
Value of a = 1
Value of b = √2
And,
First zero of polynomial (a-b) = (1 - √2)
Second zero a = 1
Third zeroes = (a+b) = (1 + √2).
☛ Zeroes of given polynomial will be (1-√2), 1 and (1 +√2) [ Ans.]
_____________________
Verification
Given equation
x³ - 3x² + x + 1 = 0
➛ x³ - x² - 2x² + 2x - x + 1 = 0
➛ x²( x -1) - 2x ( x - 1) - 1( x - 1) = 0
➛ ( x - 1) ( x² - 2x - 1) = 0
➛ ( x - 1).(x²-x-x√2-x+1+√2+x√2-√2-2) =0
➛ (x-1).[x(x-1-√2)-1(x-1-√2)+√2(x-1-√2)] = 0
➛ (x-1).[(x-1+√2)(x-1-√2)] = 0
➛ (x-1) = 0 Or, (x-1+√2) = 0 Or, (x-1-√2) = 0
➛ x = 1, Or. x = (1-√2) Or , x = (1+√2)
Hence, zeroes of this equation are 1 , (1+√2) and (1-√2)
these are same .
So, we can say that our solution is absolutely right .
__________________
⏭️⏭️@vaibhavshinde145@