CBSE BOARD X, asked by Bobbyaahana571, 9 months ago

If the zeros of the polynomial f(x)=xcube-3xsquare+x+1 are a-b a, a+b find a and b

Answers

Answered by Anonymous
15

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 .

__________________

Answered by vaibhavshinde145
3

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@

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