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For what value of `a`, does the equation ax²-(a+1)x+3=0, have roots lying between 1 and 2?

For what value of `a`, does the equation ax²-(a+1)x+3=0, have roots lying between 1 and 2?

Grade:11

4 Answers

Arun
25750 Points
6 years ago
The interval doesn’t include the boundary points 1,2. So the roots can’t be 1 or 2.
So the sum will be greater than 2 and less than 4. Similarly the product will be more than 1 and less than 4.
sum=a+1/a=1+1/a and product=3/a
2
=>1
Here the a is always positive as the sum and product of roots is positive. so we can safely multiply a on both sides in the above inequalities.
=>a
=> a belongs to the interval (1/3,1) AND a belongs to (3/4,3)
Both the conditions have to be true at the same time.
hence, a belongs to the interval (3/4,1)
Rahul Jain
10 Points
6 years ago
Just a bit of a modification there...For roots to be real D≥0a²+1+2a-12a≥0a²-10a+1≥0Therefore, a 5+2√6Also 3/4
Abu Azbar
27 Points
6 years ago
No real values of a make the roots of the equation ax^2-(a+1)x+3 between 1. and 3..So the approved answer is wrong
yathartha gupta
71 Points
6 years ago
Home » Forum » Algebra » For what value of `a`, does the equation...For what value of `a`, does the equation ax²-(a+1)x+3=0, have roots lying between 1 and 2?one month agoAnswers : (3)The interval doesn’t include the boundary points 1,2. So the roots can’t be 1 or 2.So the sum will be greater than 2 and less than 4. Similarly the product will be more than 1 and less than 4.sum=a+1/a=1+1/a and product=3/a2=>1Here the a is always positive as the sum and product of roots is positive. so we can safely multiply a on both sides in the above inequalities.=>a=> a belongs to the interval (1/3,1) AND a belongs to (3/4,3)Both the conditions have to be true at the same time.hence, a belongs to the interval (3/4,1) Ok

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