Question

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

Answer 1

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➲ ʏᴏᴜʀ ǫᴜᴇsᴛɪᴏɴ:⍰

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

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➻❥ᴀɴsᴡᴇʀ:☑

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 .

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Answer 2

✰ Qᴜēsᴛíõñ :-

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

✪ Söʟúᴛîøɴ :-

Given polynomial x³ - 3x² + x + 1

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

Therefore, 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 - b9

= 1/1

= 1

➙ a² - ab + a² + ab + a² - b² = 1

➙ 3a² - b² = 1

Here, a = 1 and b = + √2

So, 3(1)² - b² = 1

➙ 3 - b² = 1

➙ b² = 2 ➙ B = √2 = + √2