If (x – 1) and ( x – 2) are the factors of x^4 – px^2 + q, then the value of root(p+q) is
Answers
EXPLANATION.
- GIVEN
( x - 1) and ( x - 2 ) are the factors of polynomial
=> x^4 - px^2 + q = 0
Find value of ( P + q)
According to the question,
( x - 1 ) is a factors of a polynomial
=> x - 1 = 0
=> x = 1
put the value of x = 1 in equation
we get,
=> (1)^4 - P(1)^2 + q = 0
=> 1 - P + q = 0
=> 1 + q = P .......(1)
( x - 2 ) is a factors of polynomial
=> x - 2 = 0
=> x = 2
put the value of x = 2 in polynomial
we get,
=> (2)^4 - P(2)^2 + q = 0
=> 16 - 4p + q = 0 .....(1)
From equation (1) and (2) we get,
put the value of equation (1) in equation (2)
we get,
=> 16 - 4 ( 1 + q) + q = 0
=> 16 - 4 - 4q + q = 0
=> 12 - 3q = 0
=> q = 4
put the value of q = 4 in equation (1)
we get,
=> 1 + q = P
=> 1 + 4 = P
=> P = 5
Therefore,
value of P = 5 and q = 4
value of [ P + q ]
=> [ 5 + 4 ] = 9
value of root ( P + q)
=> √ P + q = √9 = 3
Answer:
9
Step-by-step explanation:
substitute x=1 and x=2 in equation to get two equations in p and q .solve for p and q you can get p=5 ,q=4.