Question

the point (2t^2 + 2t +4, t^2 + t +1) lies on the line x+2y=1 for A) all real values of t B) some real values of t C) t=-4+- 7^1/2 D) no real values of t

Answer 1

Given :  point (2t^2 + 2t +4, t^2 + t +1) lies on the line x+2y=1  

To Find :  Correct option :

A) all real values of t

B) some real values of t

C) t=-4± √7

D) no real values of t

Solution:

point (2t² + 2t +4, t² + t +1) lies on the line x+2y=1  

=> 2t² + 2t +4  + 2( t² + t +1)  = 1

=> 2t² + 2t +4  + 2 t² + 2t +2   = 1

=> 4 t² + 4t  + 5  = 0

t =( -4 ± √16 - 60 )/(2*4)

=> t  =( - 4 ±  √-44)/8

=> t  =( - 4 ±  i√44)/8

Hence Values of t are not real  

Hence no real values of t is correct answer

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Answer 2

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