Problem 118 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Solution of \(\left|x^{2}+4 x+3\right|+2 x+5=0\) is (A) 4 (B) \(-4\) (C) \(-1-\sqrt{3}\) (D) \(1+\sqrt{3}\)

Step-by-Step Solution

Answer

The solution is \(-1 - \sqrt{3}\)

Step 1: Understanding the Absolute Value

To solve the equation \(|x^2 + 4x + 3| + 2x + 5 = 0\), we start by considering the expression inside the absolute value, \(|x^2 + 4x + 3|\). The absolute value will affect the equation in cases where \(x^2 + 4x + 3\) is negative or positive, so we must consider both cases.

Step 2: Solve the Quadratic Equation

First, solve the quadratic expression \(x^2 + 4x + 3 = 0\) to find its roots. Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), where \(a = 1\), \(b = 4\), \(c = 3\). Calculate \(b^2 - 4ac = 16 - 12 = 4\). Thus, \(x = \frac{-4 \pm 2}{2}\), which gives roots \(x = -3\) and \(x = -1\).

Step 3: Consider the Intervals

The roots \(-3\) and \(-1\) divide the x-axis into three intervals: \((-\infty, -3)\), \((-3, -1)\), and \((-1, \infty)\). In each interval, we will determine the sign of \(x^2 + 4x + 3\) and consequently how the absolute value affects the equation.

Step 4: Solve for the Interval \((-\infty, -3)\)

In the interval \((-\infty, -3)\), choose a test point like \(x = -4\). Substitute into \(x^2 + 4x + 3\): \((-4)^2 + 4(-4) + 3 = 16 - 16 + 3 = 3\). Hence, it is positive. Now the equation \(|x^2 + 4x + 3| + 2x + 5 = 0\) becomes \(x^2 + 4x + 3 + 2x + 5 = 0\). Simplifying gives \(x^2 + 6x + 8 = 0\). Solving this using the quadratic formula yields no real solutions.

Step 5: Solve for the Interval \((-3, -1)\)

In the interval \((-3, -1)\), choose \(x = -2\). Substitute into \(x^2 + 4x + 3\): \((-2)^2 + 4(-2) + 3 = 4 - 8 + 3 = -1\), which is negative. So, we use \(-x^2 - 4x - 3\) instead of \(|x^2 + 4x + 3|\). The equation becomes \(-x^2 - 4x - 3 + 2x + 5 = 0\), or \(-x^2 - 2x + 2 = 0\). Solving for this gives \(x = -1 \pm \sqrt{3}\). However, only the root \(x = -1 - \sqrt{3}\) is valid in this interval.

Step 6: Solve for the Interval \((-1, \infty)\)

In the interval \((-1, \infty)\), choose \(x = 0\). Substitute into \(x^2 + 4x + 3\): \(0^2 + 4(0) + 3 = 3\), which is positive. The equation \(|x^2 + 4x + 3| + 2x + 5 = 0\) becomes \(x^2 + 4x + 3 + 2x + 5 = 0\). Simplifying gives \(x^2 + 6x + 8 = 0\), with no real solutions in this interval.

Step 7: Final Solution

The only valid solution from the discussed intervals is \(x = -1 - \sqrt{3}\), valid in the interval \((-3, -1)\). Thus, the solution to the equation \(|x^2 + 4x + 3| + 2x + 5 = 0\) is \(-1 - \sqrt{3}\).

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two. The general form is \( ax^2 + bx + c = 0 \). These equations are called quadratics because "quad" means square, and their defining feature is the \( x^2 \) term. Understanding quadratic equations is crucial since they can describe various real-life scenarios such as projectile paths and optimization problems.

• The coefficients \(a\), \(b\), and \(c\) are real numbers, with \(a eq 0\).
• The graph of a quadratic equation forms a parabola.
• Quadratic equations can have 0, 1, or 2 real roots, depending on the discriminant \(b^2 - 4ac\).

The roots or solutions of a quadratic equation are the values of \(x\) that make the equation equal to zero. We explore these solutions to understand when the quadratic expression changes from positive to negative values.

Absolute Values

Absolute values measure the distance of a number from zero on the number line, regardless of direction. The absolute value of a number \(x\) is denoted by \(|x|\), where:

• If \(x \geq 0\), then \(|x| = x\).
• If \(x < 0\), then \(|x| = -x\).

In our problem, we have an absolute value equation: \(|x^2 + 4x + 3| + 2x + 5 = 0\). This equation becomes more complex, as we need to analyze how the sign of \(x^2 + 4x + 3\) changes over different intervals.

Understanding how absolute values influence equations is essential for solving them accurately, as we must consider both the positive and negative scenarios. This often leads to solving multiple cases or sub-problems.

Interval Testing

Interval testing, or interval analysis, is a method used to evaluate expressions over specific areas of the number line. For quadratic equations, it is particularly useful when we need to determine the sign of an expression over various intervals divided by the roots.

• First, find the roots of the quadratic expression \(x^2 + 4x + 3 = 0\), which are \(-3\) and \(-1\).
• These roots divide the x-axis into intervals: \((-\infty, -3)\), \((-3, -1)\), and \((-1, \infty)\).
• In each interval, choose a test point to determine the sign of the quadratic expression.

This method helps us understand how the quadratic term affects the absolute value and thus assists in solving the equation \(|x^2 + 4x + 3| + 2x + 5 = 0\) consistently for each interval.

Quadratic Formula

The quadratic formula is a universal method for finding the roots of any quadratic equation \(ax^2 + bx + c = 0\). It is derived from the process of completing the square and provides a straightforward way to solve for \(x\).

The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• Here, \(b^2 - 4ac\) is called the discriminant.
• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root (a repeated root).
• If negative, there are no real roots.

In the original exercise, the quadratic formula was applied to \(x^2 + 4x + 3 = 0\), resulting in the roots \(-3\) and \(-1\). These roots helped us set up the intervals necessary for further analysis and ultimately solving the problem.