What Is The Solution To The Equation? Square Root -2x-5-4=x A. -7 And -3 B. 3 And 7 C. -3 D. 7
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Introduction
Mathematical equations are fundamental to understanding various concepts in algebra, calculus, and beyond. One common challenge students and math enthusiasts face is solving equations involving square roots and linear expressions. In this article, we will thoroughly analyze the equation:
\[
\sqrt{-2x - 5} - 4 = x
\]
and determine which of the provided options — A. -7 and -3, B. 3 and 7, C. -3, or D. 7 — represent its solutions.
Understanding how to approach such equations requires a systematic approach, including isolating the square root, squaring both sides, and verifying solutions to avoid extraneous roots introduced during the process.
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Understanding the Equation
The given equation is:
\[
\sqrt{-2x - 5} - 4 = x
\]
This involves a square root of a linear expression, which imposes certain restrictions:
- The expression inside the square root must be non-negative:
\[
-2x - 5 \geq 0
\]
- The solution set must satisfy this domain restriction.
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Step 1: Domain Restrictions
Before solving, let's analyze the domain:
\[
-2x - 5 \geq 0
\]
Rearranged as:
\[
-2x \geq 5
\]
Dividing both sides by -2 (note that dividing by a negative flips the inequality):
\[
x \leq -\frac{5}{2}
\]
which simplifies to:
\[
x \leq -2.5
\]
Conclusion: The solutions must satisfy \( x \leq -2.5 \).
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Step 2: Isolate the Square Root
Add 4 to both sides:
\[
\sqrt{-2x - 5} = x + 4
\]
Now, since the left side is a square root (non-negative), the right side must also be non-negative for the equality to hold:
\[
x + 4 \geq 0 \Rightarrow x \geq -4
\]
Combining with the domain restriction:
\[
-4 \leq x \leq -2.5
\]
This interval contains potential solutions; we will test values within this range or proceed algebraically.
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Step 3: Square Both Sides
Squaring both sides to eliminate the square root:
\[
(\sqrt{-2x - 5})^2 = (x + 4)^2
\]
which simplifies to:
\[
-2x - 5 = (x + 4)^2
\]
Expand the right side:
\[
-2x - 5 = x^2 + 8x + 16
\]
Bring all terms to one side to form a quadratic:
\[
0 = x^2 + 8x + 16 + 2x + 5
\]
Simplify:
\[
0 = x^2 + (8x + 2x) + (16 + 5)
\]
\[
0 = x^2 + 10x + 21
\]
---
Step 4: Solve the Quadratic Equation
The quadratic is:
\[
x^2 + 10x + 21 = 0
\]
Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 10 \), \( c = 21 \).
Calculate discriminant:
\[
\Delta = 10^2 - 4 \times 1 \times 21 = 100 - 84 = 16
\]
Compute roots:
\[
x = \frac{-10 \pm \sqrt{16}}{2} = \frac{-10 \pm 4}{2}
\]
Thus,
- For the positive root:
\[
x = \frac{-10 + 4}{2} = \frac{-6}{2} = -3
\]
- For the negative root:
\[
x = \frac{-10 - 4}{2} = \frac{-14}{2} = -7
\]
Potential solutions are: \( x = -3 \) and \( x = -7 \).
---
Step 5: Verify Solutions Against Domain and Original Equation
Recall the domain restriction:
\[
x \leq -2.5
\]
- \( x = -3 \):
- Satisfies \( x \leq -2.5 \).
- Check the original equation:
\[
\sqrt{-2(-3) - 5} - 4 = -3
\]
Calculate inside the root:
\[
-2(-3) - 5 = 6 - 5 = 1
\]
Square root:
\[
\sqrt{1} = 1
\]
Left side:
\[
1 - 4 = -3
\]
Right side:
\[
-3
\]
Both sides are equal, so \( x = -3 \) is a valid solution.
- \( x= -7 \):
- Satisfies \( x \leq -2.5 \).
- Check the original equation:
\[
\sqrt{-2(-7) - 5} - 4 = -7
\]
Inside the root:
\[
14 - 5 = 9
\]
Square root:
\[
\sqrt{9} = 3
\]
Left side:
\[
3 - 4 = -1
\]
Right side:
\[
-7
\]
Since \(-1 \neq -7\), \( x= -7 \) is an extraneous solution introduced during squaring and must be discarded.
---
Final Solution:
The only valid solution within the domain is:
\[
\boxed{x = -3}
\]
which corresponds to option C. -3.
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Summary and Key Takeaways
- When solving equations involving square roots, always determine the domain restrictions based on the radicand.
- Isolate the square root and then square both sides carefully.
- After obtaining potential solutions, verify each in the original equation to eliminate extraneous roots.
- In this problem, only \( x = -3 \) satisfies both the equation and the domain constraints.
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Additional Tips for Solving Similar Equations
- Always check the radicand: It must be non-negative for real solutions.
- Verify solutions: Squaring can introduce extraneous solutions; always substitute back into the original equation.
- Graphical methods: Plotting both sides can offer visual insight into solutions.
- Practice with different equations: Handling equations with nested radicals or multiple variables enhances problem-solving skills.
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Frequently Asked Questions (FAQs)
Q1: Why does squaring sometimes add extraneous solutions?
A1: Because squaring both sides eliminates the square root, but it can also turn false statements into true ones. Therefore, solutions obtained after squaring must be verified in the original equation.
Q2: How do I know when a radical expression is valid?
A2: The expression inside the square root must be greater than or equal to zero.
Q3: Can the solutions be complex?
A3: Yes, but in real-world problems and standard algebra, solutions are typically real numbers unless specified otherwise.
Q4: How can I improve my skills in solving radical equations?
A4: Practice diverse problems, visualize equations graphically, and always verify solutions to ensure correctness.
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Conclusion
The solution to the equation \( \sqrt{-2x - 5} - 4 = x \) is x = -3, which aligns with option C. This process illustrates the importance of domain considerations, careful algebraic manipulation, and verification of solutions. Mastering these techniques enhances problem-solving skills and deepens understanding of algebraic equations involving radicals.
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If you want to improve your math skills further, consider practicing similar problems and exploring algebraic concepts in depth. Remember, systematic approaches and verification are key to solving complex equations accurately.
