Question
If b > a , then the equation, (x - a) (x - b) - 1 = 0, has:
- both roots in [a, b]
- both roots in
- both roots in
- one root in
Hint:
The definition of a quadratic as a second-degree polynomial equation demands that at least one squared term must be included. It also goes by the name quadratic equations. Here we have to find the roots of equation, (x - a) (x - b) - 1 if b > a.
The correct answer is: one root in
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as: (x - a) (x - b) - 1 = 0. Simplifying it, we get:
x2−(a+b)x−ab−1=0
Now lets find the discriminant. The formula is:
D=b2-4ac
Applying it, we get:
D=(a+b)2−4(ab−1)
D=(a−b)2+4
D>0
So now we can say that it has two real roots. Also f(a)=−1 and f(b)=−1 but b > a which eventually means that a and b are distinct as coefficient of x2 is position (it is 1), minima of f(x) is between a and b.
Therefore, one of the roots will be in the interval of (−α,a) and the other root will be in the interval (b,α).
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, one of the roots will be in the interval of (−α,a) and the other root will be in the interval (b,α).
Related Questions to study
If 
be that roots 
where 
, such that 
and 
then the number of integral solutions of λ is
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, the number of integral solutions of λ is in between 
If be that roots where , such that and then the number of integral solutions of λ is
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The equation of the directrix of the conic 
is
The equation of the directrix of the conic is
The conic with length of latus rectum 6 and eccentricity is
The conic with length of latus rectum 6 and eccentricity is
For the circle 
centre and radius are
For the circle centre and radius are
The polar equation of the circle of radius 5 and touching the initial line at the pole is
The polar equation of the circle of radius 5 and touching the initial line at the pole is
The circle with centre at and radius 2 is
The circle with centre at and radius 2 is
Statement-I : If 
Statement-II :If 
Which of the above statements is true
Statement-I : If
Statement-II :If
Which of the above statements is true
Then
R, let [x] denote the greatest integer 
x, then value of
+
+ 
+…+
is -
