Problem 36
Question
If \(a\) and \(b\) are odd integers then \([x]^{2}+a[x]+b=0\) (where [ \(\cdot\) ] denotes greatest integer function) has (A) finite number of roots (B) infinite number of roots (C) no roots (D) None of these
Step-by-Step Solution
Verified Answer
The equation has no roots.
1Step 1: Understand the Problem
We are given a quadratic equation of the form \([x]^2 + a[x] + b = 0\), where \(a\) and \(b\) are odd integers, and \([\cdot]\) denotes the greatest integer function. We need to determine the number of roots the equation has.
2Step 2: Analyze the Quadratic Equation
Rewrite the equation as \([x]^2 + a[x] + b = 0\). Here, \([x]\) represents the greatest integer less than or equal to \(x\), which means for any integer \(x = n\), \([x] = n\). Our task is to solve \(n^2 + an + b = 0\) for integer values of \(n\).
3Step 3: Substitute Values and Solve
Since \(a\) and \(b\) are odd, substitute integer values for \(n\) in \(n^2 + an + b = 0\). Since \(a\) and \(b\) are odd, the quadratic can be rearranged to \(n(n + a) = -b\). Notice that the product \(n(n + a)\) must be even to satisfy an equality with an odd \(b\), which is impossible because \(ab\) should be odd, hence their product can never yield \(-b\).
4Step 4: Conclusion of Root Analysis
Conclude from the analysis: since an odd product can never equal \(-b\) due to parity (even*even=even, odd*odd=odd, odd*even=odd etc.) constraints, the equation \(n^2 + an + b = 0\) has no integer solutions for \(n\). Therefore, \([x]^2 + a[x] + b = 0\) has no integer roots.
Key Concepts
Quadratic EquationOdd IntegersInteger Solutions
Quadratic Equation
A quadratic equation is a second-degree polynomial equation that generally takes the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants with \( a eq 0 \). The solutions to a quadratic equation, also known as roots, are the values of \( x \) that satisfy the equation.
- The equation can have zero, one, or two real roots.
- To solve a quadratic equation, you can use various methods, such as factoring, completing the square, or the quadratic formula.
- The discriminant \( b^2 - 4ac \) helps determine the number and type of roots.
Odd Integers
Odd integers are numbers that cannot be evenly divided by two. They have a remainder of one when divided by two.
- Examples of odd integers include \(-3, 1, 5, 7,\) and many more.
- An odd number can generally be represented in the form \( 2k + 1 \), where \( k \) is an integer.
Integer Solutions
Finding integer solutions to an equation means identifying integer values that make the equation true. When working with functions that involve the greatest integer function, identifying these solutions can be more complex.
- Integer solutions are whole numbers like \(-2, 0, 3\), etc.
- These solutions are particularly relevant in problems involving polynomials because not all polynomial equations have integer solutions.
- For a quadratic equation like this exercise's, if integer solutions exist, they can be found by substitution or derived methods.
Other exercises in this chapter
Problem 34
If the roots of the equation \(a x^{2}+b x+c=0\) are of the form \(\frac{\alpha}{\alpha-1}\) and \(\frac{\alpha+1}{\alpha}\), then the value of \((a+b+c)^{2}\)
View solution Problem 35
If \(\alpha, \beta\) be roots of \(x^{2}+p x+1=0\) and \(\gamma, \delta\) be the roots of \(x^{2}+q x+1=0\), then \((\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)
View solution Problem 37
If \(\log _{9}\left(x^{2}-5 x+6\right)>\log _{3}(x-4), x\) belongs to (A) \((-\infty, 4)\) (B) \((4, \infty)\) (C) \((-\infty,-4) \cup(4, \infty)\) (D) no real
View solution Problem 38
Let \(a, b, c\) be real numbers, \(a \neq 0 .\) If \(\alpha\) is a root of \(a^{2} x^{2}+b x+c=0, \beta\) is a root of \(a^{2} x^{2}-b x-c=0\) and \(0
View solution