Hi I got this question from my tuition teacher
Given that a and b are real numbers such that 0 < a < 1 and 0< b < 1 find the probability that x^2 + sqrt(a)x + b = 0 have real roots.
Thanks for your help!
have real roots means
discriminant >= 0
a - 4b >= 0
0 < 4b < 4
-4 < a - 4b < 1
Draw number line and do by area (something like area)
So probability of a - 4b >= 0 is
area from 0 to 1 divided by area from -4 to 1
= 1 / 5
Eagle,
Not really correct.
For real roots a - 4b> = 0 ===> b <= a/4
Note that we have not made use of the fact that 0 < a < 1 and 0< b < 1.
Draw a 1 x 1 box with "b" as the vertical axis and "a" as the horizontal axis. The region satisfied by b <= a/4 will give all values that satisfy the need for real roots.
Hence the shaded area is 1/8
So the probability is 1/8
Quite a nice question.
Originally posted by greengoblin:Eagle,
Not really correct.
For real roots a - 4b> = 0 ===> b <= a/4
Note that we have not made use of the fact that 0 < a < 1 and 0< b < 1.
Draw a 1 x 1 box with "b" as the vertical axis and "a" as the horizontal axis. The region satisfied by b <= a/4 will give all values that satisfy the need for real roots.
Hence the shaded area is 1/8
So the probability is 1/8
Quite a nice question.
Hi,
Why not 1/4? Thanks!
Cheers,
Wen Shih
TS, the question will not appear in "O" level Emaths and Add Maths as it is a
combination of quadratic function, inequality and area, and probability. It can be
tested for H1 Maths.
Nevertheless, the question is fun.
---------------------------------------------------------------------------------------------------
Question
Given that a and b are real numbers such that 0 < a < 1 and 0< b < 1 find the probability that x^2 + sqrt(a)x + b = 0 have real roots.
Steps
(1) 1x^2 + sqrt(a)x + b = 0
ax^2 + bx + c = 0 (Quadratic Equation)
So, a = 1 , b = sqrt(a) , c = b
(2) For real roots, b^2 - 4ac > = 0
[sqrt(a)]^2 - 4(1)(b) > = 0
a - 4b > = 0
b < = a/4
(3) Since a and b are real numbers such that 0 < a < 1 and 0< b < 1,
draw a 1 x 1 box with "b" as the vertical axis and "a" as the horizontal axis.
The area of the box is 1 x 1 = 1 units^2
(4) Next, draw the inequality b < = a/4 in the box.
First, draw the line b = a/4.
When a = 1, b = 1/4
So, the inequality will be the area below the line b = a/4.
The required area will be the area of the triangle enclosed by the line
b = a/4 , a = 1 and b = 0
(5) Area of the triangle = 1/2 x base x height
= 1/2 x 1/4 x 1
= 1/8 units^2
(6) Hence, given that a and b are real numbers such that 0 < a < 1 and 0< b < 1
the probability that x^2 + sqrt(a)x + b = 0 have real roots
area of the triangle
= -----------------------------
area of the square box
1/8 units^2
= -------------------------
1 units^2
= 1/8
Hi,
Thanks for your well-explained solution. I forgot to consider area of triangle and kept thinking it a rectangle instead.
Some corrections:
1. b <= a/4.
2. The area is enclosed by the lines y = x/4, x = 1 and y = 0.
It is a good thinking question and suitable material for generalisation too :)
Cheers,
Wen Shih
Hi wee_ws, think Lee012lee has explained the idea in full. My bad...I always give the sketch of the proof rather than the proof itself. Will try to work on this bad habit of mine.