Question

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

Answer 1

Answer:

Step-by-step explanation:

the sum of the zeroes will be 3 as the sum always equals to = -(b)/a for any polynomial like of the form = ax3 + bx2 + cx + d,

so

a-b + a + a+b = 3

3a = 3

a = 1

also the product of the zeroes = -d/a

(a-b)(a+b)(a)= -1

(1-b)(1+b) = -1

1 - $b^{2}$ = -1

2 = $b^{2}$

b = $\sqrt{2}$

therefore, a and b is equal to 1 and $\sqrt{2}$ respectively

hope this helps

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Answer 2

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

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Given polynomial :- x³- 3x²+ x + 1

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

∴ , Sum of the zeroes

⇒ (a - b) + a + (a + b)

⇒ -(-3 ) / 1 = 3

So, 3a = 3 ⇒ a = 1

∴ Sum of the products of its zeroes taken 2 at a time.

= a (a - b) + a (a + b) + (a + b) (a - b)

= 1/1 = 1

⇒ a²- ab + a² + ab + a²- b² = 1

⇒ 3a²- b² = 1

So, 3 ( 1 )²- b² = 1

⇒ 3 - b² = 1

⇒ b² = 2 ⇒b = √2 = ± √2

Here, a = 1 and b = ± √2

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