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Answer 1

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Here is your answer,
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1.) x² - 2x + p   = 0
     3² - 2*3 + p = 0
     9 - 6 + p      = 0
     3 + p           = 0
                 p     = -3

2.) Sum of roots = (3+√2) + (3 - √2)
                           = 6
     Product of roots = (3+√2)(3-√2)
                               = 7
    quadratic polynomial = x² - 6x +7

3.) α+β = 4
      αβ = 6
     (α+β)² = 16
     α² + β² + 2αβ = 16
     α² + β² +12 = 16
     α² + β² = 4

4.) 3x² - x - 4 = 0
     3x² - 4x + 3x - 4 = 0
     x(3x-4) +1(3x-4) = 0
    (x+1)(3x-4) = 0
     x = -1         x = ⁴/₃
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Answer 2

(1)

Given f(x) = x^2 - 2x + p.

Given 3 is a zero of f(x).

Plug x = 3 in f(x), we get

= > (3)^2 - 2(3) + p = 0

= > 9 - 6 + p = 0

= > 3 + p = 0

p = -3


(2)

Given Zeroes of the polynomial are $3 + \sqrt{2} , 3 - \sqrt{2}$

Therefore the Zeroes of the Quadratic polynomial is:

= > x^2 - (sum of roots)x + (product of roots).

$= \ \textgreater \ x^2 - (3 + \sqrt{2} + 3 - \sqrt{2} )x + (3 + \sqrt{2} )(3 - \sqrt{2} )$

$= \ \textgreater \ x^2 - 6x + (3^2 - ( \sqrt{2})^2) )$

$= \ \textgreater \ x^2 - 6x + (9 - 2)$

$= \ \textgreater \ x^2 - 6x + 7$




(3).

Given 

f(x) = x^2 - 4x + 6.

It is in the form of ax^2 + bx + c, we get

a = 1, b = -4, c = 6.

Now,

We know that sum of zeroes (alpha + Beta) = -b/a

                                                                         = -(-4)/1

                                                                         = 4


We know that product of roots (Alphabeta) = c/a

                                                                        = 6/1

                                                                        = 6.


We know that alpha^2 + beta^2 = (alpha + beta)^2 - 2alphabeta

                                                      = (4)^2 - 2(6)

                                                      = 16 - 12

                                                      = 4.



(4) 

Given f(x) =  3x^2 - x - 4.

Zero of the polynomial f(x) = 0.

= > 3x^2 - x - 4 = 0

= > 3x^2 - 4x + 3x - 4 = 0

= > x(3x - 4) + 1(3x - 4) = 0

= > (3x - 4)(x + 1) = 0

x = 4/3, -1


Therefore 4/3, -1 are the zeroes of the polynomial.



Hope this helps!