Problem 9
Question
Let \(a, b, c\) be positive real numbers, such that \(b x^{2}+\) \(\left(\sqrt{(a+c)^{2}+4 b^{2}}\right) x+(a+c) \geq 0, \forall x \in R\), then \(a, b, c\) are in: (A) G.P. (B) A.P. (C) H.P. (D) None of these
Step-by-Step Solution
Verified Answer
The numbers \(a, b, c\) are in Arithmetic Progression (A.P.).
1Step 1: Analyze the given inequality
We are given the inequality \(b x^{2}+ \left(\sqrt{(a+c)^{2}+4 b^{2}}\right) x+(a+c) \geq 0\) for all real numbers \(x\). This is a quadratic equation in terms of \(x\) and for it to be non-negative for all \(x\), its discriminant must be zero or negative.
2Step 2: Calculate the discriminant
The general form of a quadratic equation \(Ax^2 + Bx + C\) has a discriminant \(\Delta = B^2 - 4AC\). Plugging the coefficients into the formula, we have \( A = b, B = \sqrt{(a+c)^2 + 4b^2}, C = (a+c) \). Thus, the discriminant becomes \( \Delta = (\sqrt{(a+c)^2 + 4b^2})^2 - 4b(a+c) \).
3Step 3: Simplify the discriminant expression
The expression \((\sqrt{(a+c)^2 + 4b^2})^2\) simplifies to \((a+c)^2 + 4b^2\). Therefore, the discriminant is \((a+c)^2 + 4b^2 - 4b(a+c)\). Simplifying further, we get \((a+c)^2 + 4b^2 - 4ab - 4bc\).
4Step 4: Set the discriminant to zero
Since the quadratic must be non-negative for all \(x\), the discriminant should be zero: \((a+c)^2 + 4b^2 - 4ab - 4bc = 0\). Expanding, we have \(a^2 + 2ac + c^2 + 4b^2 - 4ab - 4bc = 0\).
5Step 5: Rearrange terms and identify condition
Rearrange the expression: \(a^2 + c^2 + 2ac = 4ab + 4bc - 4b^2\). Notice that \(a, b, c\) are in Arithmetic Progression (A.P.) if \((a+c) = 2b\). Since we want the equation to hold as equality, substituting \(a+c = 2b\) satisfies it.
Key Concepts
DiscriminantArithmetic ProgressionPositive Real Numbers
Discriminant
In mathematics, the discriminant is a key concept when analyzing quadratic equations. Given the general quadratic form \(Ax^2 + Bx + C\), the discriminant is calculated using the formula \( \Delta = B^2 - 4AC \). The discriminant tells us about the nature of the roots the quadratic equation will have:
- If \(\Delta > 0\), the quadratic has two distinct real roots.
- If \(\Delta = 0\), there is exactly one real root, meaning the quadratic is a perfect square.
- If \(\Delta < 0\), the quadratic has no real roots, meaning its roots are complex numbers.
Arithmetic Progression
An Arithmetic Progression (A.P.) is a sequence of numbers in which the difference between consecutive terms remains constant. This difference is known as the common difference. For example, in the sequence 3, 5, 7, 9, the common difference is 2. In the context of our problem, noticing the relationship between \(a, b,\) and \(c\) helps. For them to be in an arithmetic progression, they must satisfy the condition \(2b = a + c\). This implies:
- if \(a, b, c\) are successive terms, then \(b - a = c - b\)
- The middle term \(b\) is the average of \(a\) and \(c\)
Positive Real Numbers
Positive real numbers are the set of real numbers that are greater than zero. They are fundamental in mathematics, serving numerous purposes across different areas such as analysis, geometry, and algebra. In inequalities, dealing with positive real numbers guarantees that any multiplicative inversion or logarithmic operation will yield defined results and remain meaningful. In our quadratic inequality exercise, specifying that \(a, b,\) and \(c\) are positive brings uniformity and helps regulate the expressions involved. This ensures that the solutions derived from setting up equations such as \(a + c = 2b\) make logical and mathematical sense. Working with positive real numbers gives clarity and precision, especially in forming and solving equations like quadratic forms, which depend on strict positivity to remain valid and interpretable.
Other exercises in this chapter
Problem 7
If \(a, b, c, d\) and \(p\) are distinct real numbers such that \(\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+d^{2}\right) \leq 0\)
View solution Problem 8
Let S denotes the set of all values of \(x\) for which the equation \(2 x^{2}-2(2 a+1) x+a(a+1)=0\) has one root less than \(a\) and other root greater than \(a
View solution Problem 11
If \(0 \leq x
View solution Problem 12
If one of the roots of the equation \(x^{2}-(p+1) x+p^{2}+\) \(p-8=0\) is greater than 2 and the other root is smaller than 2 , then \(p\) is such that (A) \(-\
View solution