What Are The Solutions Of The Equation $9x^4 - 2x^2 - 7 = 0$? Use $u$ Substitution To Solve.A. $x = \pm \sqrt{\frac{7}{9}}$ And $ X = ± 1 X = \pm 1 X = ± 1 [/tex]B. $x = \pm \sqrt{\frac{7}{9}}$ And $x =

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Introduction

Solving polynomial equations can be a challenging task, especially when dealing with high-degree equations. In this article, we will explore the solutions of the equation $9x^4 - 2x^2 - 7 = 0$ using the $u$ substitution method. This method involves substituting a new variable $u$ in terms of $x$ to simplify the equation and make it easier to solve.

The $u$ Substitution Method

The $u$ substitution method is a powerful technique used to solve polynomial equations. It involves substituting a new variable $u$ in terms of $x$ to simplify the equation. In this case, we will substitute $u = x^2$ into the equation $9x^4 - 2x^2 - 7 = 0$. This will give us a quadratic equation in terms of $u$, which we can then solve using the quadratic formula.

Substituting $u = x^2$ into the Equation

Let's substitute $u = x^2$ into the equation $9x^4 - 2x^2 - 7 = 0$. This gives us:

$$9u^2 - 2u - 7 = 0$$

Solving the Quadratic Equation

Now that we have a quadratic equation in terms of $u$, we can solve it using the quadratic formula. The quadratic formula is given by:

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

In this case, $a = 9$, $b = -2$, and $c = -7$. Plugging these values into the quadratic formula, we get:

$$u = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(9)(-7)}}{2(9)}$$

Simplifying this expression, we get:

$$u = \frac{2 \pm \sqrt{4 + 252}}{18}$$

$$u = \frac{2 \pm \sqrt{256}}{18}$$

$$u = \frac{2 \pm 16}{18}$$

Finding the Solutions for $u$

Now that we have the solutions for $u$, we can find the corresponding solutions for $x$. Recall that $u = x^2$, so we can take the square root of both sides to get:

$$x = \pm \sqrt{u}$$

Finding the Solutions for $x$

Now that we have the solutions for $u$, we can find the corresponding solutions for $x$. Plugging in the values of $u$, we get:

$$x = \pm \sqrt{\frac{2 + 16}{18}}$$

$$x = \pm \sqrt{\frac{18}{18}}$$

$$x = \pm \sqrt{1}$$

$$x = \pm 1$$

$$x = \pm \sqrt{\frac{2 - 16}{18}}$$

$$x = \pm \sqrt{\frac{-14}{18}}$$

$$x = \pm \sqrt{\frac{-7}{9}}$$

$$x = \pm \sqrt{\frac{7}{9}}$$

Conclusion

In this article, we used the $u$ substitution method to solve the equation $9x^4 - 2x^2 - 7 = 0$. We substituted $u = x^2$ into the equation and solved the resulting quadratic equation using the quadratic formula. We then found the corresponding solutions for $x$ by taking the square root of both sides. The solutions to the equation are $x = \pm \sqrt{\frac{7}{9}}$ and $x = \pm 1$.

Discussion

The $u$ substitution method is a powerful technique used to solve polynomial equations. It involves substituting a new variable $u$ in terms of $x$ to simplify the equation and make it easier to solve. In this case, we used the $u$ substitution method to solve the equation $9x^4 - 2x^2 - 7 = 0$. The solutions to the equation are $x = \pm \sqrt{\frac{7}{9}}$ and $x = \pm 1$.

Final Answer

The final answer is: $\boxed{x = \pm \sqrt{\frac{7}{9}} \text{ and } x = \pm 1}$

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Introduction

In our previous article, we used the $u$ substitution method to solve the equation $9x^4 - 2x^2 - 7 = 0$. In this article, we will answer some frequently asked questions about solving this equation using the $u$ substitution method.

Q: What is the $u$ substitution method?

A: The $u$ substitution method is a technique used to solve polynomial equations by substituting a new variable $u$ in terms of $x$ to simplify the equation.

Q: Why do we use the $u$ substitution method?

A: We use the $u$ substitution method to simplify the equation and make it easier to solve. By substituting $u = x^2$, we can transform the equation into a quadratic equation in terms of $u$, which is easier to solve.

Q: How do we find the solutions for $u$?

A: To find the solutions for $u$, we use the quadratic formula. The quadratic formula is given by:

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

In this case, $a = 9$, $b = -2$, and $c = -7$. Plugging these values into the quadratic formula, we get:

$$u = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(9)(-7)}}{2(9)}$$

Simplifying this expression, we get:

$$u = \frac{2 \pm \sqrt{4 + 252}}{18}$$

$$u = \frac{2 \pm \sqrt{256}}{18}$$

$$u = \frac{2 \pm 16}{18}$$

Q: How do we find the solutions for $x$?

A: To find the solutions for $x$, we take the square root of both sides of the equation $u = x^2$. This gives us:

$$x = \pm \sqrt{u}$$

Q: What are the solutions to the equation $9x^4 - 2x^2 - 7 = 0$?

A: The solutions to the equation $9x^4 - 2x^2 - 7 = 0$ are $x = \pm \sqrt{\frac{7}{9}}$ and $x = \pm 1$.

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

A: Yes, we can use other methods to solve this equation, such as factoring or using the rational root theorem. However, the $u$ substitution method is a powerful technique that can be used to solve this equation.

Q: What are some common mistakes to avoid when using the $u$ substitution method?

A: Some common mistakes to avoid when using the $u$ substitution method include:

• Not substituting the correct value of $u$ into the equation
• Not using the correct values of $a$, $b$, and $c$ in the quadratic formula
• Not taking the square root of both sides of the equation $u = x^2$

Q: How can we apply the $u$ substitution method to other polynomial equations?

A: We can apply the $u$ substitution method to other polynomial equations by substituting a new variable $u$ in terms of $x$ to simplify the equation. We can then use the quadratic formula or other methods to solve the resulting equation.

Conclusion

In this article, we answered some frequently asked questions about solving the equation $9x^4 - 2x^2 - 7 = 0$ using the $u$ substitution method. We discussed the $u$ substitution method, how to find the solutions for $u$ and $x$, and some common mistakes to avoid when using this method. We also discussed how to apply the $u$ substitution method to other polynomial equations.

Final Answer

The final answer is: $\boxed{x = \pm \sqrt{\frac{7}{9}} \text{ and } x = \pm 1}$