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Answers
Answer:
x
2
+x+k=0⇒(1)
2
+1+k=0⇒1+1+k=0⇒k=−2
(ii) 2x
2
+kx+
2
=0⇒2(1)
2
+k(1)+
2
=0⇒2+k+
2
=0⇒k=−2−
2
(iii) kx
2
−
2
x+1=0⇒k(1)
2
−
2
(1)+1=0⇒k−
2
+1=0⇒k=
2
−1
(iv) kx
2
−3x+k=0⇒k(1)
2
−3(1)+k=0⇒k−3+k=0⇒2k=−3⇒k=−
2
3
Step-by-step explanation:
Solutions:-
I) Given Polynomial P(x) = x^2-x-k
Given factor = x+1
Since x+1 is a factor of P(x) then it satisfies the given equation and by the Factor Theorem
P(-1) = 0
=> (-1)^2-(-1)-k = 0
=> 1+1-k = 0
=> 2-k = 0
=>2=k
The value of k=2
ii)
Given Polynomial P(x) = 2x^2-kx-√2
Given factor = x+1
Since x+1 is a factor of P(x) then it satisfies the given equation and by the Factor Theorem
P(-1) = 0
=>2 (-1)^2-k(-1)-√2= 0
=> 2(1)+k-√2 = 0
=>2+k-√2=0
=>k = √2-2
Or
k = √2(1-√2)
The value of k = √2 -2 or √2(1-√2)
iii)
Given Polynomial P(x) = kx^2-√2x-1
Given factor = x+1
Since x+1 is a factor of P(x) then it satisfies the given equation and by the Factor Theorem
P(-1) = 0
=> k(-1)^2-√2(-1)-1=0
=> k(1)+√2-1 = 0
=> k+√2-1 = 0
=> k = 1-√2
The value of k = 1-√2
iv)
Given Polynomial P(x) = kx^2+3x-k
Given factor = x+1
Since x+1 is a factor of P(x) then it satisfies the given equation and by the Factor Theorem
P(-1) = 0
=> k(-1)^2+3(-1)-k=0
=> k(1)-3-k = 0
=> k-3-k = 0
=> -3=0
But is is not valid
if the given polynomial is kx^2+3x+k then
We get k value
k(-1)^2+3(-1)+k=0
=> k(1)-3+k = 0
=> k-3+k = 0
=>2k -3=0
=>2k = 3
=> k = 3/2
The value of k = 3/1
Used formulae:-
Factor Theorem:-
P(x) is a polynomial of the degree greater than or equal to 1 and x-a is another linear polynomial if x-a is a factor then P(a) = 0 vice-versa.