Question

Question :-

If the zeroes of the polynomial X³-3x²+x+1 are a-b, a, a+b, Find a and b.

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Answer 1

Answer:

a−b,a,a+b are zeroes of x³ −3x² +x+1

⇒a−b+a+a+b= sum of zeroes = 3

⇒3a=3

⇒a=1

(a−b)(a)(a+b)= product of zeroes =−1

⇒(1−b)(1+b)=−1

⇒1−b² =−1

⇒b² =2

⇒b =± √2

Answer 2

Given,

p(x) = x³ - 3x² + x + 1

And

a-b, a and a-b are the zeros of p(x).

To find,

The value of a and b.

Solution,

Let,

$\alpha = a - b \\ \beta = a \\ \gamma = a + b$

We know that,

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

=> a - b + a + a+ b = 3

=> 3a = 3

=> a = 1

$\alpha \beta \gamma = \frac{ - d}{a}$

=> a-b × a × a+b = -1

=> (a-b)(a+b) ×a =- 1

=> a² - b² × a = -1 ------> 1

Putting the value of a in eq. 1

=> (1)² - b² × 1 = -1

=> 1 - b² = -1

=> -b² = -1-1

=> -b² = -2

=> b = ±√2

Hence, a = 1 and b = ±√2.

Hope it helps :)