Question

If the zeroes of polynomial x³-3x²+x+1 are (a-b),a, (a+b), find a and b.

Answer 1

Answer:

→ a = 1  and b = ±√2 .

Step-by-step explanation:

Given polynomial is f(x) = x³ - 3x² + x + 1 .

Here  a = 1 , b = -3 , c = 1 , d = 1 .

Let α = ( a - b ) , β = a and γ = ( a + b ) .

As we know,

→ α + β + γ = -b/a .

⇒ ( a - b ) + a + ( a - b ) = -(-3)/1 .

⇒ 3a = 3 .

⇒ a = 3/3 .

∴ a = 1 .

And,

→ αβ + βγ + γα = c/a .

⇒ a( a - b ) + a( a + b ) + ( a + b )( a - b ) = 1/1 .

⇒ a² - ab + a² + ab + a² - b² = 1 .

⇒ 3a² - b² = 1 .

⇒ ( 3 × 1² ) - b² = 1 .         { ∵ a = 1 }

⇒ 3 - b² = 1 .

⇒ b² = 3 - 1 .

⇒ b² = 2 .

∴ b = ±√2 .

Hence, it is solved .

Answer 2

ANSWER

Given:-

→ p(x)= x³ -3x² +x+1

and the zeroes are

→ (a-b), a ,(a+b)

$\alpha \: + \beta \: + \gamma = \frac{ - b}{a}$

Here ,

→ a = 1

→ b = -3

→ c = 1

→ d = 1

→ a - b + a + a + b = -(-3)/1

→ 3a = 3

→ a = 3/3

→ a = 1.

Now,

$\alpha \times \beta \times y = \frac{ - d}{a}$

→ (a - b) (a) (a+b) = -d/a

→ (a-b) (a+b) (a)= -1/1

→ a² - b² × a = -1

put the value of a in it

→ 1² - b² ×1 = -1

→ 1 - b² = -1

→ -b² = -1-1

→ -b² = -2

→ b = ±√2