Real Solutions For \sqrt{3x-1}=2x-5: A Math Deep Dive

Let's dive into the intriguing world of algebraic equations and figure out just how many real solutions the equation $\sqrt{3x-1}=2x-5$ truly possesses. This isn't just about crunching numbers; it's about understanding the behavior of functions and the conditions under which an equation holds true. When we talk about real solutions, we're referring to the values of $x$ that satisfy the equation and are actual numbers on the number line, not imaginary ones. The presence of a square root introduces a unique challenge, as it imposes certain restrictions on the possible values of $x$. We need to ensure that the expression inside the square root, $3x-1$, is non-negative, meaning $3x-1 \ge 0$, which simplifies to $x \ge 1/3$. This initial condition is crucial because the square root of a negative number is not a real number. Furthermore, since the square root symbol ($\sqrt{}$) conventionally denotes the principal (non-negative) square root, the right-hand side of the equation, $2x-5$, must also be non-negative. This gives us another condition: $2x-5 \ge 0$, which means $x \ge 5/2$. Combining both conditions, $x \ge 1/3$ and $x \ge 5/2$, we find that any valid solution must satisfy $x \ge 5/2$. This preliminary analysis is a vital first step in solving this type of equation, as it helps us to eliminate extraneous solutions that might arise during the solving process. Without considering these domain restrictions, we might be tempted to accept answers that, while mathematically derived, do not actually fulfill the original equation's requirements within the realm of real numbers. The graphical interpretation is also helpful here: the condition $x \ge 5/2$ tells us that we are only interested in the part of the graph of $y = \sqrt{3x-1}$ where $x$ is greater than or equal to $2.5$, and the part of the graph of $y = 2x-5$ where $x$ is greater than or equal to $2.5$. The solutions to our equation will be the $x$-coordinates of the points where these two graphs intersect within this restricted domain.

To tackle the equation $\sqrt{3x-1}=2x-5$ directly, the standard approach involves isolating the square root term and then squaring both sides to eliminate it. We've already ensured that the square root term is isolated on the left-hand side. Now, let's square both sides: $(\sqrt{3x-1})^2 = (2x-5)^2$. This operation yields $3x-1 = (2x)^2 - 2(2x)(5) + 5^2$, which simplifies to $3x-1 = 4x^2 - 20x + 25$. Our goal is to rearrange this into a standard quadratic equation form, $ax^2+bx+c=0$. To do this, we move all terms to one side: $0 = 4x^2 - 20x - 3x + 25 + 1$. Combining like terms, we arrive at the quadratic equation: $4x^2 - 23x + 26 = 0$. Now, we need to find the roots of this quadratic equation. We can use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, where $a=4$, $b=-23$, and $c=26$. First, let's calculate the discriminant, $\Delta = b^2-4ac$, which tells us about the nature of the roots. $\Delta = (-23)^2 - 4(4)(26) = 529 - 16(26) = 529 - 416 = 113$. Since the discriminant $\Delta = 113$ is positive, the quadratic equation $4x^2 - 23x + 26 = 0$ has two distinct real roots. These roots are given by $x = \frac{-(-23) \pm \sqrt{113}}{2(4)} = \frac{23 \pm \sqrt{113}}{8}$. So, the potential solutions are $x_1 = \frac{23 + \sqrt{113}}{8}$ and $x_2 = \frac{23 - \sqrt{113}}{8}$. It's important to remember that squaring both sides of an equation can sometimes introduce extraneous solutions, which are solutions that satisfy the squared equation but not the original equation. This is why we must check our potential solutions against the domain restrictions we established earlier, namely $x \ge 5/2$. Let's approximate the value of $\sqrt{113}$. Since $10^2=100$ and $11^2=121$, $\sqrt{113}$ is between 10 and 11, let's say approximately $10.6$. Then, $x_1 \approx \frac{23 + 10.6}{8} = \frac{33.6}{8} = 4.2$ and $x_2 \approx \frac{23 - 10.6}{8} = \frac{12.4}{8} = 1.55$. Comparing these approximate values with our condition $x \ge 5/2 = 2.5$: $x_1 \approx 4.2$ is greater than $2.5$, so it is a valid potential solution. However, $x_2 \approx 1.55$ is less than $2.5$, which means it violates our domain restriction and is likely an extraneous solution. Therefore, it's highly probable that only one of these roots is a true solution to the original equation.

To definitively determine the number of real solutions for the equation $\sqrt{3x-1}=2x-5$, we must rigorously check the potential solutions we found, $x_1 = \frac{23 + \sqrt{113}}{8}$ and $x_2 = \frac{23 - \sqrt{113}}{8}$, against the condition $x \ge 5/2$. As we approximated, $\sqrt{113}$ is between 10 and 11. Let's use a more precise value for clarity. $\sqrt{113} \approx 10.63$. For $x_1$: $x_1 = \frac{23 + \sqrt{113}}{8} \approx \frac{23 + 10.63}{8} = \frac{33.63}{8} \approx 4.20$. Now, let's check if $4.20 \ge 2.5$. Yes, it is. So, $x_1 = \frac{23 + \sqrt{113}}{8}$ is a valid solution. For $x_2$: $x_2 = \frac{23 - \sqrt{113}}{8} \approx \frac{23 - 10.63}{8} = \frac{12.37}{8} \approx 1.55$. Now, let's check if $1.55 \ge 2.5$. No, it is not. Therefore, $x_2 = \frac{23 - \sqrt{113}}{8}$ is an extraneous solution and must be discarded. Another way to verify this is to substitute the values back into the original equation. For $x_1$: $\sqrt{3\left(\frac{23 + \sqrt{113}}{8}\right)-1} = \sqrt{\frac{69 + 3\sqrt{113}}{8}-1} = \sqrt{\frac{69 + 3\sqrt{113} - 8}{8}} = \sqrt{\frac{61 + 3\sqrt{113}}{8}}$. The right side is $2\left(\frac{23 + \sqrt{113}}{8}\right)-5 = \frac{23 + \sqrt{113}}{4}-5 = \frac{23 + \sqrt{113} - 20}{4} = \frac{3 + \sqrt{113}}{4}$. Squaring this, we get $\left(\frac{3 + \sqrt{113}}{4}\right)^2 = \frac{9 + 6\sqrt{113} + 113}{16} = \frac{122 + 6\sqrt{113}}{16} = \frac{61 + 3\sqrt{113}}{8}$. Since $\sqrt{\frac{61 + 3\sqrt{113}}{8}} = \frac{3 + \sqrt{113}}{4}$ (both sides are positive and their squares are equal), $x_1$ is indeed a solution. For $x_2$: Since $x_2 < 5/2$, the term $2x_2 - 5$ will be negative. However, the left side, $\sqrt{3x_2-1}$, must be non-negative (as it's a principal square root). A non-negative number cannot equal a negative number, so $x_2$ cannot be a solution. This confirms that only one real solution exists for the equation $\sqrt{3x-1}=2x-5$. This process of checking for extraneous solutions is fundamental when dealing with equations involving radicals or rational expressions, as the algebraic manipulations performed can sometimes lead to solutions that don't satisfy the original constraints.

Visualizing the Solutions: A Graphical Perspective

To truly appreciate why there's only one real solution, let's consider the graphical interpretation of the equation $\sqrt{3x-1}=2x-5$. We can think of this equation as finding the intersection points of two functions: $y = \sqrt{3x-1}$ and $y = 2x-5$. The function $y = \sqrt{3x-1}$ represents the upper half of a parabola opening to the right, with its vertex at $(1/3, 0)$. Its domain is $x \ge 1/3$, and its range is $y \ge 0$. The function $y = 2x-5$ represents a straight line with a slope of 2 and a y-intercept of -5. As established earlier, for a solution to exist, we must have $x \ge 5/2$ (or $x \ge 2.5$). This is because the output of the square root function is always non-negative, so $2x-5$ must also be non-negative. Let's analyze the behavior of these two graphs for $x \ge 2.5$. At $x=2.5$, the value of $y = \sqrt{3x-1}$ is $\sqrt{3(2.5)-1} = \sqrt{7.5-1} = \sqrt{6.5} \approx 2.55$. The value of $y = 2x-5$ at $x=2.5$ is $2(2.5)-5 = 5-5 = 0$. So, at $x=2.5$, the square root function is above the line. Now consider the rates at which these functions increase. The derivative of $y = \sqrt{3x-1}$ is $y' = \frac{1}{2\sqrt{3x-1}} \cdot 3 = \frac{3}{2\sqrt{3x-1}}$. The derivative of $y = 2x-5$ is simply 2. For $x > 2.5$, let's see how these slopes compare. When $x=2.5$, the slope of the square root function is $\frac{3}{2\sqrt{6.5}} \approx \frac{3}{2(2.55)} \approx \frac{3}{5.1} \approx 0.59$. This slope is less than 2, the slope of the line. As $x$ increases, the denominator $2\sqrt{3x-1}$ increases, so the slope of the square root function decreases. This means that the line $y=2x-5$, which starts below the curve $y=\sqrt{3x-1}$ at $x=2.5$ (in fact, the line is at $y=0$ and the curve is at $y=\sqrt{6.5}$), will eventually catch up to and intersect the curve because its slope is consistently greater than the slope of the curve for all $x > 1/3$. Since the line starts lower at $x=2.5$ and grows faster, it will intersect the curve exactly once for $x > 2.5$. The extraneous solution $x_2 = \frac{23 - \sqrt{113}}{8} \approx 1.55$ corresponds to an intersection point of the parabola $y^2 = 3x-1$ and the line $y=2x-5$, but it's on the lower half of the parabola ($y < 0$), which is not represented by $y = \sqrt{3x-1}$. The graphical analysis perfectly complements the algebraic method, confirming that there is indeed only one point of intersection within the valid domain, thus yielding a single real solution.

Conclusion: The Uniqueness of the Solution

In conclusion, after meticulously analyzing the equation $\sqrt{3x-1}=2x-5$, we have found that it possesses exactly one real solution. This conclusion was reached through a combination of algebraic manipulation and domain restriction analysis. We first established the necessary conditions for a real solution to exist: $x \ge 1/3$ from the square root and $x \ge 5/2$ from the result of the square root, leading to a combined requirement of $x \ge 5/2$. Squaring both sides led to the quadratic equation $4x^2 - 23x + 26 = 0$, which yielded two potential real roots: $x_1 = \frac{23 + \sqrt{113}}{8}$ and $x_2 = \frac{23 - \sqrt{113}}{8}$. Upon checking these potential solutions against the domain restriction $x \ge 5/2$, we determined that $x_1 \approx 4.20$ satisfies the condition, while $x_2 \approx 1.55$ does not. Therefore, $x_2$ is an extraneous solution. The graphical interpretation further solidified this finding, illustrating that the line $y=2x-5$ intersects the curve $y=\sqrt{3x-1}$ at precisely one point within the valid domain. This thorough examination ensures that we have accounted for all possibilities and arrived at the correct number of solutions. The process highlights the importance of checking for extraneous solutions when solving radical equations.

For further exploration into solving radical equations and understanding extraneous solutions, you can refer to resources from Math is Fun or the Khan Academy.