Analyzing the Setup
The equation provided is ∣x2+4x+3∣+∣2x+5∣=0.
The modulus function is a one-way mirror; for any real number a, the modulus ∣a∣ is always greater than or equal to zero. This is the core geometric reality we must visualize.
We are adding two quantities, ∣x2+4x+3∣ and ∣2x+5∣, and their sum is zero. Since both terms are non-negative, the only way their sum can be zero is if each term is zero individually.
This is the 'AND' condition. We are looking for a value that forces both terms to vanish simultaneously.
The Quadratic Investigation
Let us break this down. First, we focus on the term ∣x2+4x+3∣=0.
This implies that
x2+4x+3=0. We can factorize it by splitting the middle term:
(x+3)(x+1)=0
This gives us two potential candidates for x: x=−3 and x=−1. These are the only two points on the number line where the first part of our equation is satisfied.
The Linear Constraint
Now, we turn our attention to the second term: ∣2x+5∣=0.
This implies
2x+5=0. Solving for
x, we get
2x=−5, which means:
x=−25=−2.5
This is the only point on the number line where the second part of our equation is satisfied. If x is anything else, this term will be strictly positive.
The Intersection Verdict
Here is where the magic happens. We have two sets of solutions: the first set is {−3,−1} and the second set is {−2.5}.
For the original equation to hold, we need a value of
x that exists in both sets. We are looking for the intersection:
{−3,−1}∩{−2.5}=∅
As we look at these values on the number line, we see they are distinct. There is no overlap.
Because there is no single value of x that can satisfy both conditions at the same time, we must conclude that there is no solution.