Question

Find the zeroes of the polynomial such that thier sum and product are 1/4 and -1

Answer 1

Answer:

Zeros : (1 + √65)/8 , (1 - √65)/8

Polynomial : x² - x/4 - 1

Note:

★ (a + b)² = a² + 2ab + b²

★ (a - b)² = a² - 2ab + b²

★ a² - b² = (a + b)(a - b)

★ (a + b)³ = a³ + b³ + 3ab(a + b)

★ (a - b)³ = a³ + b³ - 3ab(a - b)

★ a³ + b³ = (a + b)(a² - ab + b²)

★ a³ - b³ = (a - b)(a² + ab + b²)

Solution:

Let A and B be the zeros of the polynomial .

Also,

It is given that , the sum and the product of zeros of the polynomial are 1/4 and -1 respectively .

Thus,

A + B = 1/4 ---------(1)

AB = -1 ---------(2)

Now,

=> (A - B)² = (A + B)² - 4AB

=> (A - B)² = (1/4)² - 4(-1)

=> (A - B)² = 1/16 + 4

=> (A - B)² = (1 + 64)/16

=> (A - B)² = 65/16

=> A - B = √(65/16)

=> A - B = √65 / 4 ----------(3)

Now,

Adding eq-(1) and (2) , we get ;

=> A + B + A - B = 1/4 + √65/4

=> 2A = (√65 + 1)/4

=> A = (√65 + 1)/8

Now ,

Putting A = (√65 + 1)/8 in eq-(1) , we get ;

=> A + B = 1/4

=> (√65 + 1)/8 + B = 1/4

=> B = 1/4 - (√65 + 1)/8

=> B = (2 - √65 - 1)/8

=> B = (1 - √65)/8

Hence,

The zeros of the polynomial are ;

(1 + √65)/8 , (1 - √65)/8

Moreover,

The polynomial will be given as ;

x² - (A + B)x + A•B

=> x² - (1/4)x + (-1)

=> x² - x/4 - 1