find the value of k for which x=1 is a root of the equation ( x square + k x + 3 ) = 0 . Also, find the other root.
Answers
Answered by
116
Hii friend,
X = 1
P(X) = X²+KX-3
Putting X = 1
P(1) = (1)² + K × 1 +3
=> 1 + K + 3 = 0
=> K = -4
P(X) = X²-4X+3
=> X²-3X-X+3
=> X(X-3) -1(X-3)
=> (X-3) (X-1)
=> X-3 = 0. OR X-1 = 0
=> X = 3. OR X = 1
Hence,
the Value of K is -4 and 3 , 1 are the two zeros of the polynomial X²-4X+3.
HOPE IT WILL HELP YOU... :-)
X = 1
P(X) = X²+KX-3
Putting X = 1
P(1) = (1)² + K × 1 +3
=> 1 + K + 3 = 0
=> K = -4
P(X) = X²-4X+3
=> X²-3X-X+3
=> X(X-3) -1(X-3)
=> (X-3) (X-1)
=> X-3 = 0. OR X-1 = 0
=> X = 3. OR X = 1
Hence,
the Value of K is -4 and 3 , 1 are the two zeros of the polynomial X²-4X+3.
HOPE IT WILL HELP YOU... :-)
Answered by
20
Answer:
Hii friend,
X = 1
P(X) = X²+KX-3
Putting X = 1
P(1) = (1)² + K × 1 +3
=> 1 + K + 3 = 0
=> K = -4
P(X) = X²-4X+3
=> X²-3X-X+3
=> X(X-3) -1(X-3)
=> (X-3) (X-1)
=> X-3 = 0. OR X-1 = 0
=> X = 3. OR X = 1
Hence,
the Value of K is -4 and 3 , 1 are the two zeros of the polynomial X²-4X+3.
HOPE IT WILL HELP YOU... :-)
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