341For 2 I guess you have omitted mentioning that a,b,c are real (in which case Option D is right
Q1: No. of ordered pair (x,y) satisfying the Eqn x2+1=y and y2+a=x is:
Q2: If the Eqn ax2+bx+c=0 and x3+3x2+3x+2=0 has two common roots, then:
A) a = b (Not Equal to) c
B) a (Not equal to) b = c
C) a=b=c
D) a=-b=c
Please Explain!
-
UP 0 DOWN 0 1 4
4 Answers
341
71I think it is complete.... However, Can you explain a bit to me. How did you reach upto the solution or Whenever such type of questions are asked how to approach!
341The cubic equation can be written as (x+1)^3=-1 which has roots -1, \omega, \omega^2
Now, if a,b,c are real, the roots of the quadratic are obviously \omega, \omega^2 (as they are conjugates)
Hence the quadratic is a(x^2-x+1) from which the conclusion follows.
71Trying One?? I think It has no ordered pair of Solutions? Am I right?