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Chapter 3: Problem 143

Find all real and imaginary solutions to \(x^{5}+13 x^{3}+36 x=0\)

Short Answer

Expert verified
The real solution is \(x = 0\). The imaginary solutions are \(x = \pm 2i\) and \(x = \pm 3i\).

Step by step solution

01

Factor Out the Greatest Common Factor

Take the greatest common factor out of the equation. For the equation \(x^{5}+13 x^{3}+36 x=0\), the greatest common factor is \(x\). So, factor \(x\) out to get: \(x(x^{4} + 13x^{2} + 36) = 0\).
02

Set Each Factor to Zero

Set each factor from the factored equation \(x(x^{4} + 13x^{2} + 36) = 0\) to zero. This gives us two simpler equations: \(x = 0\) and \(x^{4} + 13x^{2} + 36 = 0\).
03

Solve the Linear Equation

Solve the linear equation \(x = 0\). This gives the real solution \(x = 0\).
04

Use Substitution for the Quartic Equation

For the equation \(x^{4} + 13x^{2} + 36 = 0\), use the substitution \(u = x^2\). This turns the quartic equation into a quadratic one: \(u^2 + 13u + 36 = 0\).
05

Solve the Quadratic Equation

Solve the quadratic equation \(u^2 + 13u + 36 = 0\) using the quadratic formula \(u = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\). Here, \(a = 1\), \(b = 13\), and \(c = 36\). Therefore: \(u = \frac{-13 \pm \sqrt{13^2 - 4(1)(36)}}{2(1)} = \frac{-13 \pm \sqrt{169 - 144}}{2} = \frac{-13 \pm \sqrt{25}}{2} = \frac{-13 \pm 5}{2}\). This gives two solutions: \(u = -4\) and \(u = -9\).
06

Find the Solutions for x

Substitute back \(u = x^2\). So we have \(x^2 = -4\) and \(x^2 = -9\). Solving these, we get imaginary solutions: \(x = \pm 2i\) and \(x = \pm 3i\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

factoring polynomials
Factoring polynomials is fundamental in solving polynomial equations. In the given exercise, the polynomial equation to solve is \(x^5 + 13x^3 + 36x = 0\). The first step is to find the greatest common factor (GCF). The GCF is the largest polynomial that evenly divides each term of the equation. In our case, it鈥檚 \(x\). We factor \(x\) out to get \(x(x^4 + 13x^2 + 36) = 0\).
quartic equations
A quartic equation is a polynomial equation of degree 4. In the factored form from the previous step, we have \(x(x^4 + 13x^2 + 36) = 0\). To solve for \(x\), we need to find the roots of the quartic equation \(x^4 + 13x^2 + 36 = 0\). However, solving quartic equations directly can sometimes be challenging. Instead, we use a method called quadratic substitution.
quadratic substitution
Quadratic substitution simplifies quartic equations by changing variables. Here, we use \(u = x^2\). This converts our quartic equation \(x^4 + 13x^2 + 36 = 0\) into a quadratic equation: \(u^2 + 13u + 36 = 0\). We solve the quadratic equation using the quadratic formula: \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Plugging in \(a = 1\), \(b = 13\), and \(c = 36\), we get \( u = \frac{-13 \pm \sqrt{169 - 144}}{2} = \frac{-13 \pm 5}{2} \). Solving this, we find \(u = -4\) and \(u = -9\). Substituting back \(u = x^2\), we get \(x^2 = -4\) and \(x^2 = -9\), leading to the imaginary solutions \(x = \pm 2i\) and \(x = \pm 3i\).

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