Problem 2
Question
If \(a, b, c\) are positive real numbers, then the number of real roots of the equation \(a x^{2}+b|x|+c=0\) is (A) 0 (B) 2 (C) 4 (D) None of these
Step-by-Step Solution
Verified Answer
The equation can have 4 real roots.
1Step 1: Understand the Equation
The given equation is \(a x^2 + b |x| + c = 0\), where \(a, b, c\) are positive real numbers. We need to find the number of real roots of this equation.
2Step 2: Split the Equation Based on |x|
Since we have an absolute value, consider two cases. Case (i): \(x \geq 0\), where \(|x| = x\); the equation becomes \(a x^2 + bx + c = 0\). Case (ii): \(x < 0\), where \(|x| = -x\); the equation becomes \(a x^2 - bx + c = 0\).
3Step 3: Analyze Case for \(x \geq 0\)
For \(x \geq 0\), solve \(a x^2 + bx + c = 0\). The discriminant \(\Delta_1 = b^2 - 4ac\). For real roots, \(\Delta_1\) must be non-negative, \(\Delta_1 \geq 0\).
4Step 4: Analyze Case for \(x < 0\)
For \(x < 0\), solve \(a x^2 - bx + c = 0\). The discriminant \(\Delta_2 = b^2 - 4ac\) (same as the previous). For real roots, \(\Delta_2\) must be non-negative, \(\Delta_2 \geq 0\).
5Step 5: Conclusion on Real Roots
In both cases, the discriminant does not change and must be non-negative for real roots. If \(\Delta_1 = \Delta_2 > 0\), each quadratic equation can have 2 distinct roots, leading to a total of 4 roots for both cases combined. Therefore, there can be 4 real roots as a possibility.
Key Concepts
Absolute Value EquationsReal RootsDiscriminants
Absolute Value Equations
When dealing with quadratic equations that involve absolute values, like \(a x^2 + b|x| + c = 0\), there are a few important considerations to make. The absolute value \(|x|\) introduces a major characteristic of the equation, as it can affect the algebraic behavior based on the sign of \(x\). To address this:
- Consider two main cases: when \(x \geq 0\) and when \(x < 0\).
- For \(x \geq 0\), \(|x| = x\) and the equation can be expressed as \(a x^2 + bx + c = 0\).
- For \(x < 0\), \(|x| = -x\) and the equation translates to \(a x^2 - bx + c = 0\).
Real Roots
In quadratic equations, real roots are the solutions for which the quadratic equation holds true and they are real numbers. When solving an equation like \(a x^2 + b|x| + c = 0\), determining the number of real roots involves analyzing the two cases we've established.
- For \(x \geq 0\), solve the equation \(a x^2 + bx + c = 0\). The roots here are real if the discriminant (notated as \(\Delta_1\)) is non-negative.
- For \(x < 0\), solve \(a x^2 - bx + c = 0\). The roots are real if the discriminant (\(\Delta_2\)) is likewise non-negative.
Discriminants
The concept of the discriminant is pivotal in understanding the roots of quadratic equations. It is denoted by \(\Delta\) and typically calculated as \(b^2 - 4ac\). Here’s how it works:
- If \(\Delta > 0\), the quadratic equation has two distinct real roots.
- If \(\Delta = 0\), there is exactly one real root (a repeated root).
- If \(\Delta < 0\), the quadratic equation does not have any real roots (the roots are complex).
Other exercises in this chapter
Problem 1
The roots of the equation \(2^{x+2} \cdot 3^{\frac{3 x}{x-1}}=9\) are given by (A) \(\log _{2},\left(\frac{2}{3}\right)-2\) (B) \(3,-3\) (C) \(-2,1-\frac{\log 3
View solution Problem 3
If \(x^{2}-x+1=0\), then value of \(x^{3 n}\) is (A) 0 (B) \(-1\) (C) 1 (D) \(-1,1\)
View solution Problem 4
The number of negative integral solutions of \(x^{2} \cdot 2^{x+1}+2^{|x-3|+2}=x^{2} \cdot 2^{(3 x-3 \mid+4)}+2^{x-1}\) is (A) 4 (B) 2 (C) 1 (D) 0
View solution Problem 5
If \(\alpha\) and \(\beta(\alpha
View solution