pls answer it correctly I am poor at maths pls i will mark Brainlist
Answers
Step-by-step explanation:
1.
Given quadratic polynomial is
P(x) = x^2-(k-6) x +(2k+1)
On Comparing this with the standard quadratic Polynomial ax^2+bx+c
a = 1
b=-(k-6)
c=(2k+1)
Given that
α and β are the zeroes of P(x) then
Sum of the zeroes (α + β) = -b/a
α + β = -(-k-6)/1
α + β = k-6--------(1)
Product of the zeroes (αβ) = c/a
αβ = (2k+1)/1
αβ = 2k+1 --------(2)
Given that
α + β = αβ
From (1) & (2)
k-6 = 2k+1
=> -6-1 = 2k-k
=> -7 = k
=> k = -7
The value of k = -7
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2.
Given quadratic polynomial is
P(x) = x^4 -6x^3-26x^2+138x-35
Given zeroes = 2±√3
We know that by factor theorem
If 2+√3 and 2-√3 are zeroes then
[x-(2+√3)] and [x-(2-√3)] are the factors of P(x)
and The product of zeroes is also a factor of P(x)
=> [x-(2+√3)] [x-(2-√3)]
=> [(x-2-√3)(x-2+√3)]
=> (x-2)^2-(√3)^2
Since (a+b)(a-b)=a^2-b^2
Where a = x-2 and b=√3
=> x^2-4x+4-3
=> x^2-4x+1
Now to get the other zeroes we have to divide the given P(x) by x^2-4x+1
On dividing then
we get the quotient x^2-2x-35
To get the zeores we write x^2-2x-35 = 0
=> x^2-7x+5x-35=0
=> x(x-7)+5(x+7)=0
=> (x-7)(x+5) =0
=> x-7 = 0 or x+5 = 0
=> x=7 and x=-5
The other zeroes are 7 and -5
See the above attachment for division
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3.See the above attachment
4.
Given Quadratic Polynomial is
P(x) = x^2-5x+6
To get zeroes we write P(x) = 0
x^2-5x+6 = 0
=> x^2-3x-2x+6 = 0
=> x(x-3)-2(x-3) = 0
=> (x-3)(x-2)=0
=> x-3 = 0 or x-2 = 0
=> x=3 and 2
α and β are the zeroes of P(x) then
α = 3 and β = 2
=>1/ α =1/3
1/β =1/ 2
We know that α and β are the zeroes then the Quadratic Polynomial is K[x^2-(α+β )x+αβ ]
=> K[x^2-{(1/3)+(1/2)}x+(1/2)(1/3)]
=>K[x^2-(5/6)x+(1/6)]
=>K[6x^2-5x+1]/6
If K=6 then the Polynomial is x^2-5x+1
The required Polynomial is x^2-5x+1
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