If a and b are positive numbers and each of the quations x2+ax+2b = 0 and x2+2bx+a = 0 has real roots, then the smallest possible value of (a+b) is
1. 3
2. 4
3. 5
4. 6
62a2≤8b
and b2≤a
Now what you can do is to draw the graph of x2≤8y and y2≤x
Then draw the line x+y and find which of the parallel lines falls where it is smallest....
It reminds me of the worst chapter i have studied :P (or atleast i felt that when i studied it :P) linear programming :D
1(1) - (2) gives ...
(x - 1)(a - 2b) = 0
Either x = 1 ... or a = 2b
Now putting x = 1 in (1) ... a/b = -2 ... which is not possible as a and b are both +ve ...
Hence a = 2b is possible ...
Now for(1) ... a2 - 8b ≥ 0
Putting a = 2b in the expr. .... b≥0 and b≥2 ...
Now b ≠0 ... therefore min. value of b possible is 2
Now putting the value of b in the discriminant of (2); i.e.;
4b2 - 4a ≥ 0
min. value of a = 2 ...
Hence min. value of (a + b) is 2 + 2 = 4 = Opt. 2
341We need a^2 \ge 8b and b^2 \ge a to be simultaneously satisfied with a,b>0.
Eliminating b, we get a^3 \ge 64 \Rightarrow a \ge 4. Henceb^2 \ge a \ge 4 \Rightarrow b \ge 2.
(4,2) satisfies the given conditions and hence the min value of a+b is 6
1Sir ... I have subtracted x2 + ax + 2b = 0 .... (1) and x2 + 2bx + a = 0 ... (2) ... Isn't that possible ????
341doing that would have meaning if they had common roots for example
1Oh ya ... I got it now .... I made a mistake .... !!!
Sorry ...
Thanks NISHANT and PROPHET sir ...
1yes its answer is 6. thogh my way is diff. but answer is 6. is it right surbhi.
