Problem 53 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \(4 q^{4}-13 q^{2}+9=0\)

Step-by-Step Solution

Answer

The solutions are \(q = \frac{3}{2}, -\frac{3}{2}, 1, -1\).

Step 1: Substitute variable

Let us use a substitution to simplify the equation. Set \(u = q^2\). This transforms the original equation into a quadratic equation: \(4u^2 - 13u + 9 = 0\).

Step 2: Solve the quadratic equation

We can solve the quadratic equation \(4u^2 - 13u + 9 = 0\) using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 4\), \(b = -13\), and \(c = 9\). Plugging in these values, we get : \[u = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 4 \cdot 9}}{2 \cdot 4}\] \[u = \frac{13 \pm \sqrt{169 - 144}}{8}\] \[u = \frac{13 \pm 5}{8}\]. So, \(u = \frac{18}{8} = \frac{9}{4}\) or \(u = \frac{8}{8} = 1\). Hence, the solutions for \(u\) are \(u = \frac{9}{4}\) and \(u = 1\).

Step 3: Substitute back to find q

Since \(u = q^2\), we substitute back to find \(q\). If \(u = \frac{9}{4}\), then \(q^2 = \frac{9}{4}\) giving two solutions: \(q = \frac{3}{2}\) and \(q = -\frac{3}{2}\). Similarly, if \(u = 1\), then \(q^2 = 1\) giving solutions: \(q = 1\) and \(q = -1\). Hence, the solutions for \(q\) are \(q = \frac{3}{2}, -\frac{3}{2}, 1, -1\).

Step 4: Check the solutions

To verify the solutions, substitute each back into the original equation \(4q^4 - 13q^2 + 9 = 0\). For \(q = \frac{3}{2}\): \(4(\frac{3}{2})^4 - 13(\frac{3}{2})^2 + 9 = 0\) \[4 \cdot \frac{81}{16} - 13 \cdot \frac{9}{4} + 9 = 0\] \[ \frac{81}{4} - \frac{117}{4} + 9 = 0\] \[ 0 = 0 \] (Verified)

Key Concepts

Polynomial Equations

Polynomial equations are expressions that involve variables raised to integer powers and coefficients. They can appear in different forms and degrees, with the highest power indicating the degree of the polynomial. For example, in the equation given \(4q^4 - 13q^2 + 9 = 0\), it is a fourth-degree polynomial because the highest power of the variable, q, is 4. Solving polynomial equations can sometimes be straightforward, but often they require methods like factoring, using the quadratic formula, or substitution when they are more complex.

Substitution Method

The substitution method helps in simplifying complex polynomial equations by replacing variables with simpler expressions. For instance, in the given problem, we set \(u = q^2\). This turns the complex polynomial \(4q^4 - 13q^2 + 9 = 0\) into a more manageable quadratic equation \(4u^2 - 13u + 9 = 0\). By substituting back after solving this simpler equation, we can find the solutions for the original variable. This method is particularly useful when we can identify a part of the equation that behaves like a standard polynomial form.

Quadratic Formula

The quadratic formula is used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. In our exercise, after substituting \(u = q^2\), we got \(4u^2 - 13u + 9 = 0\). By identifying \(a = 4\), \(b = -13\), and \(c = 9\), we plugged these into the quadratic formula to obtain the solutions \[u = \frac{13 \pm 5}{8}\]. This resulted in \(u = \frac{9}{4}\) and \(u = 1\). Using the quadratic formula ensures that we can find reliable solutions for any quadratic equation if it doesn't factorize easily.

Solution Verification

Verifying solutions to an equation is crucial to ensure they are correct. After solving \(4q^4 - 13q^2 + 9 = 0\) and finding \(q = \frac{3}{2}, -\frac{3}{2}, 1, -1\), we need to substitute each back into the original equation to check them. For example, substituting \(q = \frac{3}{2}\) into the equation, we perform: \[4(\frac{3}{2})^4 - 13(\frac{3}{2})^2 + 9 = 0\]. Simplifying this, we see that the left-hand side equals zero, confirming our solution. This step ensures that what we computed is accurate and satisfies the original equation, a critical step in solving any mathematical problem.