Solving Equations By Factoring: A Step-by-Step Guide

Hey guys! Today, we're diving into a cool math problem that involves solving an equation by factoring. Specifically, we're going to tackle the equation $7 x^{1 / 2}-6 x^{5 / 2}=0$. We'll make sure we don't include any extraneous solutions, and we'll use a clever substitution to make things easier. So, grab your pencils, and let's get started!

Understanding the Problem

Before we jump into the solution, let's break down what we're dealing with. The equation $7 x^{1 / 2}-6 x^{5 / 2}=0$ might look a bit intimidating at first, with those fractional exponents. But don't worry, we'll simplify it using factoring and a handy substitution. Factoring is a method to express the equation into a product of its factors that equals zero. Then we can solve for the values of $x$ that make each factor equal to zero. This method is pretty useful for solving polynomial equations and can be extended to equations with fractional exponents like this one.

The extraneous solutions are those results that arise from the solution process, but they don't satisfy the original equation. These can appear when we manipulate the equation in ways that introduce new potential solutions that are not valid. In this case, we need to be extra careful when dealing with exponents and substitutions. Always remember to check our solutions in the original equation to weed out any imposters!

Lastly, we’ll use the substitution $u = x^{1/2}$ to simplify the equation and make it easier to work with. This substitution transforms the equation into a more manageable form, allowing us to factor it more easily. After finding the values of $u$, we’ll need to remember to substitute back to find the values of $x$.

Step-by-Step Solution

Step 1: Factoring out the Common Term

The first thing we notice in the equation $7 x^{1 / 2}-6 x^{5 / 2}=0$ is that both terms contain a power of $x$. Specifically, they both have at least $x^{1/2}$. Factoring out the common term is a crucial step that simplifies the equation and sets us up for success. We can factor out $x^{1/2}$ from both terms:

$x^{1 / 2}(7 - 6x^{5/2 - 1/2}) = 0$

Simplifying the exponent in the parenthesis, we get:

$x^{1 / 2}(7 - 6x^{4/2}) = 0$

Which further simplifies to:

$x^{1 / 2}(7 - 6x^2) = 0$

Step 2: Applying the Zero Product Property

Now that we have factored the equation into the form $x^{1 / 2}(7 - 6x^2) = 0$, we can apply the zero product property. This property states that if the product of two factors is zero, then at least one of the factors must be zero. In other words, if $AB = 0$, then either $A = 0$ or $B = 0$ (or both). Applying this to our equation, we get two possibilities:

1. $x^{1 / 2} = 0$
2. $7 - 6x^2 = 0$

Step 3: Solving for x in the First Factor

For the first factor, $x^{1 / 2} = 0$, we need to solve for $x$. Recall that $x^{1/2}$ is the same as $\sqrt{x}$. So, we have:

$\sqrt{x} = 0$

To solve for $x$, we square both sides of the equation:

$(\sqrt{x})^2 = 0^2$

This simplifies to:

$x = 0$

So, one solution is $x = 0$.

Step 4: Solving for x in the Second Factor

Now, let's tackle the second factor, $7 - 6x^2 = 0$. We want to isolate $x^2$ first:

$6x^2 = 7$

Divide both sides by 6:

$x^2 = \frac{7}{6}$

To solve for $x$, we take the square root of both sides. Remember to consider both the positive and negative square roots:

$x = \pm \sqrt{\frac{7}{6}}$

So, we have two more potential solutions: $x = \sqrt{\frac{7}{6}}$ and $x = -\sqrt{\frac{7}{6}}$.

Step 5: Checking for Extraneous Solutions

We have three potential solutions: $x = 0$, $x = \sqrt{\frac{7}{6}}$, and $x = -\sqrt{\frac{7}{6}}$. Now, we need to check each one in the original equation $7 x^{1 / 2}-6 x^{5 / 2}=0$ to make sure they are valid and not extraneous.

Checking $x = 0$:

Substitute $x = 0$ into the original equation:

$7(0)^{1 / 2} - 6(0)^{5 / 2} = 0$

$7(0) - 6(0) = 0$

$0 = 0$

So, $x = 0$ is a valid solution.

Checking $x = \sqrt{\frac{7}{6}}$:

Substitute $x = \sqrt{\frac{7}{6}}$ into the original equation:

$7(\sqrt{\frac{7}{6}})^{1 / 2} - 6(\sqrt{\frac{7}{6}})^{5 / 2} = 0$

This is a bit tricky to simplify directly, but let's rewrite $x$ as $x = \sqrt{\frac{7}{6}}$ implies $x^2 = \frac{7}{6}$. Thus $x = (7/6)^{1/2}$.

Now the equation becomes:

$7(\frac{7}{6})^{1/4} - 6(\frac{7}{6})^{5/4} = 0$

Factor out $(\frac{7}{6})^{1/4}$:

$(\frac{7}{6})^{1/4} [7 - 6(\frac{7}{6})] = 0$

$(\frac{7}{6})^{1/4} [7 - 7] = 0$

$(\frac{7}{6})^{1/4} [0] = 0$

$0 = 0$

So, $x = \sqrt{\frac{7}{6}}$ is also a valid solution.

Checking $x = -\sqrt{\frac{7}{6}}$:

Substitute $x = -\sqrt{\frac{7}{6}}$ into the original equation. Since we have $x^{1/2}$ in the original equation, and $x^{1/2} = \sqrt{x}$, we cannot take the square root of a negative number within the real number system. Therefore, $x = -\sqrt{\frac{7}{6}}$ is an extraneous solution.

Step 6: State the Solutions

After checking for extraneous solutions, we found that $x = 0$ and $x = \sqrt{\frac{7}{6}}$ are the only valid solutions.

Using the Substitution $u = x^{1/2}$

As requested, let’s see how using the substitution $u = x^{1/2}$ can help us solve this equation. We can rewrite the original equation $7 x^{1 / 2}-6 x^{5 / 2}=0$ in terms of $u$.

Since $u = x^{1/2}$, then $x = u^2$. Therefore, $x^{5/2} = (x^{1/2})^5 = u^5$. Substituting these into the original equation, we get:

$7u - 6u^5 = 0$

Now, factor out the common term $u$:

$u(7 - 6u^4) = 0$

Applying the zero product property, we have two possibilities:

1. $u = 0$
2. $7 - 6u^4 = 0$

From the first factor, $u = 0$. Since $u = x^{1/2}$, we have $x^{1/2} = 0$, which implies $x = 0$.

From the second factor, $7 - 6u^4 = 0$, we solve for $u^4$:

$6u^4 = 7$

$u^4 = \frac{7}{6}$

Taking the fourth root of both sides:

$u = \pm \sqrt[4]{\frac{7}{6}}$

Now, substitute back $u = x^{1/2}$:

$x^{1/2} = \pm \sqrt[4]{\frac{7}{6}}$

Squaring both sides:

$x = (\pm \sqrt[4]{\frac{7}{6}})^2$

$x = \sqrt{\frac{7}{6}}$

Note that squaring the positive and negative values yields the same result since $(\sqrt[4]{\frac{7}{6}})^2 = (-\sqrt[4]{\frac{7}{6}})^2$. Thus, the negative root does not lead to an additional solution. Again, we have to check for extraneous solutions.

Therefore we have $x = 0$ and $x = \sqrt{\frac{7}{6}}$ as valid solutions, and we discard $x = -\sqrt{\frac{7}{6}}$ as extraneous.

Conclusion

Alright, guys, we've successfully solved the equation $7 x^{1 / 2}-6 x^{5 / 2}=0$ by factoring, checking for extraneous solutions, and even using the substitution $u = x^{1/2}$. Remember, factoring is all about simplifying and breaking down complex equations into manageable parts. Always be mindful of extraneous solutions, especially when dealing with radicals and fractional exponents. Keep practicing, and you'll become a factoring pro in no time!