Two numbers b and c are chosen at random with replacement from the numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9. The probability that $x^2 + bx+c>0$ for all x ∈ R is
Answer & explanation
Correct answer: option 3
Since b and c both can assume values from 1 to 9.
So, total numbers pf ways of choosing b and c is 9 × 9 = 81.
Now,
$x^2 + bx + c > 0 $ for all x ∈ R
⇒ Disc < 0 i.e $b^2 - 4c < 0.$
The following table shows the possible values of b and c for which $b^2 - 4c < 0.$
|
C |
b |
Total |
|
1 2 3 4 5 6 7 8 9 |
1 1,2 1,2,3 1,2,3 1,2,3,4 1,2,3,4 1,2,3,4,5 1,2,3,4,5 1,2,3,4,5 |
1 2 3 3 4 4 5 5 5 |
|
|
|
32 |
So, favourable number of ways = 32.
Hence, required probability =$ \frac{32}{81}.$