Question

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Answer 1

Step-by-step explanation:

Solutions:-

1)

Given zeroes are 3+√5 and 3-√5

Let α = 3+√5 and β = 3-√5

We know that

The Quadratic Polynomial whose zeroes are α and β is K[x^2-(α+β)x+αβ]

On Substituting these values in the above formula

=> K[x^2-(3+√5+3-√5)x+(3+√5)(3-√5)]

=> K[x^2-(3+3)x+3^2-(√5)^2]

=> K[x^2-6x+9-5]

=>K[x^2-6x+4]

If K = 1 then the required Polynomial is x^2-6x+4

2)

Given Polynomial is P(x) = 3x+6

To get the Zero of P(x) we write P(x) = 0

=> 3x+6 = 0

=> 3x = -6

=> x = -6/3

=> x = -2

The zero of the Polynomial 3x+6 is -2

3) Number of zeroes of the given graph is 1

Since the graph touches the X - axis at only one point.

4)

Given Polynomial is x^2-6x+2

On Comparing this with the standard quadratic Polynomial ax^2+bx+c

a = 1

b=-6

c=2

We know that

Sum of the zeroes (α +β) = -b/a

=> (α +β) = -(-6)/1

(α +β) = 6 ------(1)

Product of the zeroes (αβ) = c/a

=> αβ = 2/1

αβ = 2 ---------(2)

Now

The value of 2(α+β)+3αβ

From (1)&(2)

=> 2(α+β)+3αβ

=> 2(6)+3(2)

=> 12+6

=> 18

The value of 2(α+β)+3αβ is 18