Question

.If sum of zeroes of a polynomial α+β=-8 and product of zeroes αβ=6 then form a polynomial whose zeroes are (α-β)and(α+β).

Answer 1

${x}^{2} - ( \alpha + \beta ) + \alpha \times \beta \\ {x}^{2} + (8 + \sqrt{40} )x + 8 \sqrt{40}$

Answer 2

Given : The sum of zeroes α + β = -8  and the product is αβ= 6

To Find : A polynomial whose zeroes are (α-β)and(α+β).

Solution:

α + β = -8  

αβ= 6

(α-β)² = (α+β) ² - 4αβ

=> (α-β)² = (-8)² - 4(6)

=>  (α-β)² = 64 - 24

=>  (α-β)² = 40

=> α-β =  ±2√10

A polynomial whose zeroes are (α-β)and(α+β).

Sum of zeroes

= (α-β) + (α+β).

=   ±2√10 +( -8 )

= - 8 ±2√10

product of zeroes =  (α-β)   (α+β)

=  ±2√10(8)

= ±16√10

polynomial  is :

x² - ( - 8 ±2√10)x    ±16√10

= 10x²+(8 -2√10) x  + 16  

or

10x²+(8 +2√10) x - 16  

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