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!
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!
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4 Answers
341For 2 I guess you have omitted mentioning that a,b,c are real (in which case Option D is right
71341The 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?