Question

If p and q are zeroes of polynomial x^2 + 2x - 1, find p - q ( x^2 means x square)



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Answer 1

Answer:

2√2

Step-by-step explanation:

We know, polynomials written in form of x^2 - Sx + P refer S as sum of their roots and P as product of their roots.

 Which, here, in pol. x^2 + 2x - 1

sum of roots = p + q = - 2

product of roots = pq = - 1

 Square on both sides of p + q

⇒ ( p + q )^2 = ( - 2 )^2

⇒ p^2 + q^2 + 2pq = 4

⇒ p^2 + q^2 + 2( - 1 ) = 4

⇒ p^2 + q^2 - 2 = 4

⇒ p^2 + q^2 = 4 + 2 = 6

   Subtract 2pq from both sides

⇒ p^2 + q^2 - 2pq = 6 - 2pq

⇒ ( p - q )^2 = 6 - 2( - 1 )

⇒ ( p - q )^2 = 6 + 2 = 8

⇒ p - q = √8

⇒ p - q = 2√2

Answer 2

Given : p and q are zeroes of polynomial  x² + 2x - 1  

To find :  p - q

Solution:

p and q are zeroes of polynomial x² + 2x - 1

Sum of roots  p + q  =  -2/1  = - 2

Product of roots  pq  = - 1/1 = -1

p +  q   =  - 2

Squaring both sides

=> p² + q² + 2pq  =  4

=> p² + q²  + 2(-1)  = 4

=> p² + q²   = 6

Subtracting   2pq on both sides

=> p² + q²  -2pq  = 6    -2pq

=> ( p - q)²   = 6  + 2

=>  ( p - q)²   =  8

=> p - q  = ± 2√2

Another methods

Roots p , q  are  =   (-2  ± √4 + 4)/2   =  -1 ± √2

Case 1  p = -1 + √2   & q  = -1   - √2

=> p - q = 2√2

Case 2  p = -1 -√2   & q  = -1  +√2

=> p - q = -2√2

p - q  = ± 2√2

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