A Student Stated That There Are Two Solutions To The Equation $\sqrt{3x+7}=x-1$.They Are $x=-1$ And $x=6$.Do You Agree With The Student? Explain Why Or Why Not.A. I Do Not Agree With The Student Because There Is No Real

A Critical Analysis of the Equation $\sqrt{3x+7}=x-1$

In mathematics, solving equations is a fundamental concept that requires careful analysis and attention to detail. A student recently claimed that there are two solutions to the equation $\sqrt{3x+7}=x-1$, which are $x=-1$ and $x=6$. However, upon closer examination, we will demonstrate that the student's assertion is incorrect. In this article, we will delve into the world of algebra and explore the properties of square roots to determine the validity of the student's claim.

Before we proceed, it is essential to understand the properties of square roots. The square root of a number $a$, denoted by $\sqrt{a}$, is a value that, when multiplied by itself, gives the original number $a$. In other words, $\sqrt{a} \times \sqrt{a} = a$. This definition is crucial in solving equations involving square roots.

To solve the equation $\sqrt{3x+7}=x-1$, we need to isolate the variable $x$. One way to do this is by squaring both sides of the equation. This will eliminate the square root and allow us to solve for $x$.

$\sqrt{3x+7}=x-1$

Squaring both sides:

$3x+7=(x-1)^2$

Expanding the right-hand side:

$3x+7=x^2-2x+1$

Rearranging the terms:

$x^2-5x-6=0$

Now we have a quadratic equation in the form $ax^2+bx+c=0$, where $a=1$, $b=-5$, and $c=-6$. We can solve this equation using the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substituting the values of $a$, $b$, and $c$:

$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-6)}}{2(1)}$

Simplifying:

$x=\frac{5\pm\sqrt{25+24}}{2}$

$x=\frac{5\pm\sqrt{49}}{2}$

$x=\frac{5\pm7}{2}$

Now we have two possible solutions:

$x_1=\frac{5+7}{2}=6$

$x_2=\frac{5-7}{2}=-1$

At first glance, it may seem that the student's claim is correct, as both $x=6$ and $x=-1$ satisfy the original equation. However, we must consider the domain of the square root function. The square root of a number is defined only for non-negative numbers. In other words, $\sqrt{a}$ is only defined when $a\geq0$.

Let's examine the solutions in the context of the original equation:

$\sqrt{3x+7}=x-1$

For $x=6$:

$\sqrt{3(6)+7}=(6)-1$

$\sqrt{25}=5$

This solution is valid, as the square root of 25 is indeed 5.

For $x=-1$:

$\sqrt{3(-1)+7}=(-1)-1$

$\sqrt{-2}=-2$

This solution is not valid, as the square root of a negative number is not defined in the real number system.

In conclusion, while the student's claim that there are two solutions to the equation $\sqrt{3x+7}=x-1$ may seem plausible at first, it is actually incorrect. The solution $x=-1$ is not valid, as it results in a negative number under the square root, which is not defined in the real number system. Therefore, we must agree with the student that there is only one valid solution to the equation, which is $x=6$.
A Critical Analysis of the Equation $\sqrt{3x+7}=x-1$

In our previous article, we analyzed the equation $\sqrt{3x+7}=x-1$ and demonstrated that there is only one valid solution, which is $x=6$. However, we received several questions from readers regarding the equation and its solutions. In this article, we will address some of the most frequently asked questions and provide additional insights into the equation.

Q: Why is the solution $x=-1$ not valid?

A: The solution $x=-1$ is not valid because it results in a negative number under the square root, which is not defined in the real number system. When we substitute $x=-1$ into the original equation, we get:

$\sqrt{3(-1)+7}=(-1)-1$

$\sqrt{-2}=-2$

As we mentioned earlier, the square root of a negative number is not defined in the real number system. Therefore, the solution $x=-1$ is not valid.

Q: Can we use complex numbers to solve the equation?

A: Yes, we can use complex numbers to solve the equation. Complex numbers are numbers that have both real and imaginary parts. In this case, we can rewrite the equation as:

$\sqrt{3x+7}=x-1$

Squaring both sides:

$3x+7=(x-1)^2$

Expanding the right-hand side:

$3x+7=x^2-2x+1$

Rearranging the terms:

$x^2-5x-6=0$

Now we have a quadratic equation in the form $ax^2+bx+c=0$, where $a=1$, $b=-5$, and $c=-6$. We can solve this equation using the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substituting the values of $a$, $b$, and $c$:

$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-6)}}{2(1)}$

Simplifying:

$x=\frac{5\pm\sqrt{25+24}}{2}$

$x=\frac{5\pm\sqrt{49}}{2}$

$x=\frac{5\pm7i}{2}$

Now we have two complex solutions:

$x_1=\frac{5+7i}{2}$

$x_2=\frac{5-7i}{2}$

Q: What is the significance of the complex solutions?

A: The complex solutions $x_1=\frac{5+7i}{2}$ and $x_2=\frac{5-7i}{2}$ are significant because they provide a complete solution to the equation. In other words, they represent all possible values of $x$ that satisfy the equation. The complex solutions are not just a mathematical curiosity; they have practical applications in fields such as engineering and physics.

Q: Can we use other methods to solve the equation?

A: Yes, we can use other methods to solve the equation. One such method is to use the method of substitution. We can substitute $y=x-1$ into the original equation and solve for $y$. This will give us a new equation in terms of $y$, which we can then solve.

Q: What are some real-world applications of the equation?

A: The equation $\sqrt{3x+7}=x-1$ has several real-world applications. For example, it can be used to model population growth, where the square root term represents the growth rate and the linear term represents the initial population. It can also be used to model electrical circuits, where the square root term represents the voltage and the linear term represents the current.

In conclusion, the equation $\sqrt{3x+7}=x-1$ is a complex equation that has both real and complex solutions. While the real solution $x=6$ is valid, the complex solutions $x_1=\frac{5+7i}{2}$ and $x_2=\frac{5-7i}{2}$ provide a complete solution to the equation. We hope that this Q&A article has provided additional insights into the equation and its solutions.