Question

If α and β are the zeroes of the polynomial 2x² - x – 3 , then write the polynomial whose

zeroes are α², β².

Answer should be 4x² - 13x + 9​

Answer 1

Step-by-step explanation:

α+ β = 1/2

α β = -3/2

α² + β² = (α+ β)² + 2αβ

= 1/4 - 3

= -11/4

α²β² = (αβ) ²

= 9/4

Equation = x² +11/4x +9/4

= 4x² + 11x + 9

Answer 2

Answer:

Step-by-step explanation:

Given Polynomial is 2x² - x – 3,

a=2

b= -1

c=3

Now,

α+β= -b/a

⇒ α+β= -(-1)/2

⇒ α+β= 1/2

αβ= c/a

⇒ αβ= 3/2

Now,

α²+β²= (α+β)²- 2αβ

⇒ α²+β²= (1/2)²- 2(3/2)

⇒ α²+β²= 1/4 - 3 (solve 1/4 - 3)

⇒ α²+β²= 13/4

To find a quadratic polynomial we have;

x²- (sum of zeroes)x + product of zeroes

since it is given that zeroes are α² and β²,

⇒ x²-(α²+β²)x+ (α²β²)⇒x²-(α²+β²)x+ (αβ)²

⇒  x²-(13/4 )x+ (3/2)²

⇒x²- 13/4x + 9/4

⇒ $\frac{4x^{2}-13x+9 }{4}$    { by solving x²- 13/4x + 9/4}

Therefore the required polynomial whose zeroes are α² and β² is 4x²-13x+9.