Solve $x^4 - 5x^2 - 14 = 0$ By Factoring

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Let's dive into solving polynomial equations, specifically the one you've presented: $x^4 - 5x^2 - 14 = 0$. This type of equation, where the exponents are even and follow a pattern like $x^{2n}$, $x^n$, and a constant, can often be tackled using a clever technique called factoring. It's like finding the hidden building blocks of the equation that, when multiplied together, give you the original expression. The goal here is to find the values of $x$ that make this equation true. We'll break it down step-by-step to make it super clear.

Understanding the Structure: A Quadratic in Disguise

When we first look at $x^4 - 5x^2 - 14 = 0$, it might seem a bit intimidating because of the $x^4$ term. However, notice the relationship between the exponents: 4, 2, and 0 (for the constant term). This structure is a key indicator that we can treat it as a quadratic equation in disguise. If we let a new variable, say $u$, represent $x^2$, then $u^2$ would be $(x^2)^2 = x^4$. This substitution transforms our original equation into a much more familiar form: $u^2 - 5u - 14 = 0$. This is a standard quadratic equation that we can solve for $u$ using factoring, completing the square, or the quadratic formula. For this particular problem, factoring is the most straightforward method, and it’s explicitly requested.

Factoring the Quadratic in 'u'

Now that we have $u^2 - 5u - 14 = 0$, we need to find two numbers that multiply to -14 and add up to -5. Let's think about the factors of -14: (1, -14), (-1, 14), (2, -7), and (-2, 7). We need the pair that sums to -5. If we try (2, -7), their product is $2 imes (-7) = -14$, and their sum is $2 + (-7) = -5$. Bingo! These are our numbers.

So, we can factor the quadratic equation in $u$ as $(u + 2)(u - 7) = 0$. For this product to be zero, at least one of the factors must be zero. This gives us two possibilities:

1. $u + 2 = 0
2. u - 7 = 0$

Solving for $u$ in each case, we get:

1. $u = -2$
2. $u = 7$

These are the solutions for $u$. But remember, our original problem was to solve for $x$, not $u$. So, we need to substitute back $x^2$ for $u$. This is where we find the actual roots of the original equation.

Substituting Back and Finding 'x'

We found that $u = -2$ and $u = 7$. Now, let's substitute $x^2$ back in for $u$:

Case 1: $u = -2$

Substituting $x^2$ for $u$, we get $x^2 = -2$. To solve for $x$, we take the square root of both sides. Remember that when you take the square root of both sides of an equation, you must consider both the positive and negative roots. So, $x = pm{- 2}$. Since we have a negative number under the square root, we introduce the imaginary unit, $i$, where $i = pm{-1}$. Therefore, $x = pm{2i}$ and $x = - pm{2i}$, which can be written more compactly as $x = pm{2i}$.

Case 2: $u = 7$

Substituting $x^2$ for $u$, we get $x^2 = 7$. Again, taking the square root of both sides, we consider both positive and negative roots: $x = pm{7}$. Since 7 is a positive number, the roots are real: $x = pm{7}$ and $x = - pm{7}$, which can be written as $x = pm{7}$.

Consolidating the Solutions

By combining the solutions from both cases, we have found all the roots for the original equation $x^4 - 5x^2 - 14 = 0$. The solutions are $x = pm{2i}$ and $x = pm{7}$.

Let's compare these solutions with the options provided:

A. $x= pm{7}$ and $x= pm{2}$ B. $x= pm{7i}$ and $x= pm{2i}$ C. $x= pm{7i}$ and $x= pm{2}$ D. $x= pm{7}$ and $x= pm{2i}$

Our calculated solutions are $x = pm{7}$ and $x = pm{2i}$. This perfectly matches option D.

Verifying the Solutions (Optional but Recommended)

To be absolutely sure, we can plug these values back into the original equation. Let's test $x = pm{7}$.

$x^2 = ( pm{7})^2 = 7$ $x^4 = ( pm{7})^4 = (x^2)^2 = 7^2 = 49$

Substituting into $x^4 - 5x^2 - 14 = 0$: $49 - 5(7) - 14 = 49 - 35 - 14 = 14 - 14 = 0$. This works!

Now let's test $x = pm{2i}$.

$x^2 = ( pm{2i})^2 = ( pm{4} pm{i^2}) = 4(-1) = -4$ $x^4 = (x^2)^2 = (-4)^2 = 16$

Substituting into $x^4 - 5x^2 - 14 = 0$: $16 - 5(-4) - 14 = 16 + 20 - 14 = 36 - 14 = 22$. Hmm, there seems to be a slight discrepancy here in my manual check. Let's recheck the calculation for $x^2 = -2$ and $x^2 = 7$. Ah, I see the error in my verification step for the imaginary roots. Let's re-evaluate $x^2$ and $x^4$ for $x = pm{2i}$.

If $x = pm{2i}$, then $x^2 = ( pm{2i})^2 = pm{4} pm{i^2} = 4(-1) = -4$. This is correct. If $x = pm{2i}$, then $x^4 = (x^2)^2 = (-4)^2 = 16$. This is also correct.

Substituting into $x^4 - 5x^2 - 14 = 0$: $16 - 5(-4) - 14 = 16 + 20 - 14 = 36 - 14 = 22$. The value 22 is not 0. Let me re-examine the initial factoring.

The equation is $u^2 - 5u - 14 = 0$. The factors of -14 that sum to -5 are indeed +2 and -7. So, $(u+2)(u-7)=0$. This means $u=-2$ or $u=7$.

Substituting back $u=x^2$:

Case 1: $x^2 = -2$. This gives $x = pm{-2} = pm{2i}$. And $x=- pm{2i}$. Case 2: $x^2 = 7$. This gives $x = pm{7}$. And $x=- pm{7}$.

The solutions are $x = pm{2i}, x = - pm{2i}, x = pm{7}, x = - pm{7}$.

Now let's re-evaluate the options and my check.

Option D is $x= pm{7}$ and $x= pm{2i}$. This implies that the solutions are ONLY these two values, which is not correct for a fourth-degree polynomial that should have four roots (counting multiplicity and complex roots).

Let me re-read the question and options carefully. The question asks "What are the solutions... Use factoring to solve." The options provided are pairs of solutions, not the complete set of solutions.

Let's re-check the factoring of the original equation $x^4 - 5x^2 - 14 = 0$. Let $y = x^2$. Then $y^2 - 5y - 14 = 0$. Factoring this gives $(y-7)(y+2) = 0$. So $y=7$ or $y=-2$.

Substituting back $x^2$ for $y$:

If $y=7$, then $x^2=7$, which means $x = pm{7}$ and $x = - pm{7}$. If $y=-2$, then $x^2=-2$, which means $x = pm{-2} = pm{2i}$ and $x = - pm{2i}$.

The complete set of solutions is $x = pm{7}, x = - pm{7}, x = pm{2i}, x = - pm{2i}$.

Now let's look at the options again:

A. $x= pm{7}$ and $x= pm{2}$ B. $x= pm{7i}$ and $x= pm{2i}$ C. $x= pm{7i}$ and $x= pm{2}$ D. $x= pm{7}$ and $x= pm{2i}$

Option D lists $x = pm{7}$ and $x = pm{2i}$. These are indeed two of the solutions we found. The question asks "What are the solutions", and provides options that are pairs of solutions. It seems the question or the options are not perfectly formed, as a fourth-degree polynomial typically has four roots. However, among the given choices, Option D contains two valid roots that we derived.

Let's verify option D's components:

• $x = pm{7}$: We verified this, $x^2 = 7$, $x^4 = 49$. $49 - 5(7) - 14 = 49 - 35 - 14 = 0$. Correct.
• $x = pm{2i}$: We found $x^2 = -2$, $x^4 = 4$. $4 - 5(-2) - 14 = 4 + 10 - 14 = 14 - 14 = 0$. Correct.

My previous verification for $x = pm{2i}$ had an error. Let's trace that error: When $x = pm{2i}$, $x^2 = ( pm{2i})^2 = pm{4} imes pm{i^2} = 4 imes (-1) = -4$. This part was correct. Then $x^4 = (x^2)^2 = (-4)^2 = 16$. This part was also correct. Substituting into the original equation: $x^4 - 5x^2 - 14 = 0$. So, $16 - 5(-4) - 14 = 16 + 20 - 14 = 36 - 14 = 22$. This is where the error was. Let me re-evaluate $x^2$ for $u=-2$. If $u = -2$, then $x^2 = -2$. The roots are $x = pm{-2} = pm{2i}$ and $x = - pm{2i}$.

Let's check $x = pm{2i}$ again: $x^2 = ( pm{2i})^2 = 4i^2 = -4$. This is correct. $x^4 = (x^2)^2 = (-4)^2 = 16$. This is correct.

Let's re-substitute into $x^4 - 5x^2 - 14 = 0$: $16 - 5(-4) - 14 = 16 + 20 - 14 = 36 - 14 = 22$. There is still a persistent error in my verification or understanding. Let me go back to the substitution $u=x^2$. The equation in $u$ is $u^2 - 5u - 14 = 0$. Factoring gives $(u-7)(u+2)=0$. Thus $u=7$ or $u=-2$. These are the correct values for $u$. Now, $u=x^2$. So $x^2=7$ or $x^2=-2$. These are the correct relationships.

If $x^2=7$, then $x = pm{7}$. Correct. If $x^2=-2$, then $x = pm{-2} = pm{2i}$. Correct.

Let's re-test $x = pm{2i}$ by plugging $x^2=-2$ and $x^4=4$ into the original equation. $x^4 - 5x^2 - 14 = (x^2)^2 - 5(x^2) - 14 = (-2)^2 - 5(-2) - 14 = 4 + 10 - 14 = 14 - 14 = 0$. Aha! The error was in calculating $x^4$ from $x = pm{2i}$ previously. It should be $x^4 = (x^2)^2 = (-2)^2 = 4$, not 16.

So, the solutions $x= pm{7}$ and $x= pm{2i}$ are indeed valid solutions.

Conclusion

By recognizing the equation $x^4 - 5x^2 - 14 = 0$ as a quadratic in disguise, we used substitution ($u=x^2$) to transform it into $u^2 - 5u - 14 = 0$. We then factored this quadratic into $(u-7)(u+2)=0$, yielding $u=7$ and $u=-2$. Substituting back $x^2$ for $u$, we got $x^2=7$ and $x^2=-2$. Solving these for $x$ gives us the full set of solutions: $x = pm{7}, x = - pm{7}, x = pm{2i}, x = - pm{2i}$.

Looking at the provided options, option D, $x= pm{7}$ and $x= pm{2i}$, lists two of these correct solutions. Assuming the question asks to identify a pair of correct solutions from the choices, Option D is the correct answer.

For further reading on solving polynomial equations and factoring techniques, you can explore resources like Khan Academy's algebra section or Paul's Online Math Notes. These sites offer comprehensive explanations and practice problems to deepen your understanding.