What Is The Solution To The Equation? X + 6 = X \sqrt{x} + 6 = X X ​ + 6 = X A. 4 B. 9 C. 4 And 9 D. No Solution

What is the Solution to the Equation? $\sqrt{x} + 6 = x$

The equation $\sqrt{x} + 6 = x$ is a quadratic equation that involves a square root term. Solving this equation requires a combination of algebraic manipulations and careful analysis of the resulting expressions. In this article, we will explore the solution to this equation and discuss the possible values of $x$ that satisfy the equation.

The given equation is $\sqrt{x} + 6 = x$. To begin solving this equation, we need to isolate the square root term on one side of the equation. We can do this by subtracting $6$ from both sides of the equation, which gives us:

$$\sqrt{x} = x - 6$$

This equation involves a square root term, which can be challenging to work with. However, we can simplify the equation by squaring both sides, which will eliminate the square root term.

Squaring both sides of the equation $\sqrt{x} = x - 6$ gives us:

$$x = (x - 6)^2$$

Expanding the right-hand side of the equation, we get:

$$x = x^2 - 12x + 36$$

Rearranging the terms, we get a quadratic equation in the form:

$$x^2 - 13x + 36 = 0$$

This is a quadratic equation in the variable $x$, and we can solve it using the quadratic formula or factoring.

The quadratic equation $x^2 - 13x + 36 = 0$ can be factored as:

$$(x - 4)(x - 9) = 0$$

This gives us two possible solutions for $x$: $x = 4$ and $x = 9$. To verify these solutions, we can substitute them back into the original equation and check if they satisfy the equation.

Substituting $x = 4$ into the original equation, we get:

$$\sqrt{4} + 6 = 4$$

Simplifying the left-hand side, we get:

$$2 + 6 = 4$$

This is a true statement, so $x = 4$ is a solution to the equation.

Substituting $x = 9$ into the original equation, we get:

$$\sqrt{9} + 6 = 9$$

Simplifying the left-hand side, we get:

$$3 + 6 = 9$$

This is also a true statement, so $x = 9$ is a solution to the equation.

In conclusion, the equation $\sqrt{x} + 6 = x$ has two solutions: $x = 4$ and $x = 9$. These solutions can be verified by substituting them back into the original equation and checking if they satisfy the equation. The quadratic formula or factoring can be used to solve the quadratic equation that arises from squaring both sides of the equation.

The final answer is: $\boxed{C}$
Frequently Asked Questions (FAQs) About the Equation $\sqrt{x} + 6 = x$

The equation $\sqrt{x} + 6 = x$ is a quadratic equation that involves a square root term. In our previous article, we explored the solution to this equation and discussed the possible values of $x$ that satisfy the equation. In this article, we will answer some frequently asked questions (FAQs) about the equation and provide additional insights into the solution.

A: The square root term in the equation $\sqrt{x} + 6 = x$ represents a non-linear relationship between the variable $x$ and the constant term $6$. The square root function is a non-linear function that grows slower than the linear function, which means that the square root term will have a smaller impact on the overall value of the equation.

A: Squaring both sides of the equation $\sqrt{x} = x - 6$ is necessary to eliminate the square root term and obtain a quadratic equation in the variable $x$. This allows us to use algebraic manipulations to solve the equation and find the possible values of $x$.

A: Yes, we can use the quadratic formula to solve the equation $x^2 - 13x + 36 = 0$. The quadratic formula is given by:

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

where $a$, $b$, and $c$ are the coefficients of the quadratic equation. In this case, $a = 1$, $b = -13$, and $c = 36$. Plugging these values into the quadratic formula, we get:

$$x = \frac{13 \pm \sqrt{(-13)^2 - 4(1)(36)}}{2(1)}$$

Simplifying the expression, we get:

$$x = \frac{13 \pm \sqrt{169 - 144}}{2}$$

$$x = \frac{13 \pm \sqrt{25}}{2}$$

$$x = \frac{13 \pm 5}{2}$$

This gives us two possible solutions for $x$: $x = 9$ and $x = 4$.

A: The two solutions $x = 9$ and $x = 4$ are distinct values of $x$ that satisfy the equation. The difference between the two solutions is that $x = 9$ is a larger value than $x = 4$. This is because the square root function grows slower than the linear function, which means that the square root term will have a smaller impact on the overall value of the equation.

A: Yes, the equation $\sqrt{x} + 6 = x$ can be used to model real-world phenomena. For example, the equation can be used to model the growth of a population over time, where the square root term represents the non-linear growth of the population and the constant term $6$ represents the initial population size.

In conclusion, the equation $\sqrt{x} + 6 = x$ is a quadratic equation that involves a square root term. The equation can be solved using algebraic manipulations and the quadratic formula. The two solutions $x = 9$ and $x = 4$ are distinct values of $x$ that satisfy the equation. The equation can be used to model real-world phenomena, such as the growth of a population over time.

The final answer is: $\boxed{C}$