Consider the cubic equation given by x3+ax2+bx+c = 0 , where a, b, c are real numbers. Which of the following is correct :
A. If a2-2b < 0, then the equation has one real and two imaginary roots
B. If a2-2b ≥ 0, then the equation has all real roots
C. If a2-2b > 0, then the equation has all real and distinct roots
D. If 4a3-27b2 >0 , then the equation has real and distinct roots.
Don't no why but I think that question is somewhere wrong.I may be wrong though.
21NO, its perfectly correct...
f(x)= x^3+ax^2+bx+c, f'(x) = 3x^2 + 2ax + b
discriminant of f'(x) is D=4(a^2 - 3b)
f(x)= 0, ofcourse has one real root,since its a cubic...
If f'(x)≠0 for all real x, then f(x) has no more root.. so we want D<0, i.e a^2 - 3b<0
Note that if a^2 - 2b<0 , b must be >0, so a^2 - 3b<-b<0
so option A is correct..
11For 3 real roots necessary and sufficient condition is that:
1) f'(x) should have 2 real roots. ( say a and b)
2) f(a)*f(b) <0.
For 1 real root necessary and sufficient condition is that:
Case 1: f'(x) has 2 real roots and f(a)*f(b) >0
Case 2: f'(x) has no real roots i.e the function is either increasing or decreasing always and thus it must cut x-axis at some pt thus f(x) has 1 root.
11@Shubhodip: I could'nt understand ur last argument:
How a^2-3b<0 implies a^2-2b<0 ??
71Main bhi last step tak pahucha tha.. Wahi reasoning mein gadbad hui. Khair I have marked A only..
71@Sagnik,
a2 < 3b ,for two imaginary roots, Certainly 2b < 3b since b > 0 So, a2<2b will also satisfy the given condition.
@Subhodip.
How was the exam. How much are you expecting.
11Well in my 1st post ignore condition for 1 real root. A cubic eqtn will always have a real root. Sorry for the wrong post..
341One more way of seeing this is - suppose all three roots are real, then we would have
(r_1+r_2+r_3)^2 \ge 3 (r_1r_2+r_2r_3+r_3r_1)
so that we have a^2 \ge 3 b
So if a^2 \le 2b all three arent real. Since atleast one root is real, there is exactly one real root.