Problem 78 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$4 x^{4}-8 x^{2}+3=0$$

Step-by-Step Solution

Answer

The solutions are \( x = \pm \frac{\sqrt{6}}{2} \) and \( x = \pm \frac{\sqrt{2}}{2} \).

Step 1: Identify substitution

Let a new variable, say, \(y\), equal \(x^2\). Thus, we have \(y = x^2\). Rewrite the equation in terms of \(y\):

Step 2: Substitute and simplify

Substitute \(x^2 = y\) into the original equation to get a quadratic equation: \[ 4y^2 - 8y + 3 = 0 \]

Step 3: Solve the quadratic equation

Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \(a = 4\), \(b = -8\), and \(c = 3\). Plug these values into the formula to solve for \(y\): \( y = \frac{8 \pm \sqrt{64 - 48}}{8} \) Simplify under the square root and continue: \( y = \frac{8 \pm \sqrt{16}}{8} \) \( y = \frac{8 \pm 4}{8} \) Solving these gives two solutions: \( y = \frac{8+4}{8} = \frac{12}{8} = \frac{3}{2} \) and \( y = \frac{8-4}{8} = \frac{4}{8} = \frac{1}{2} \)

Step 4: Back-substitute to find \(x\)

Recall \( y = x^2 \). Thus, set the solutions for \( y \) equal to \( x^2 \): \( x^2 = \frac{3}{2} \) and \( x^2 = \frac{1}{2} \). Take the square root of both sides for each solution: \( x = \pm \sqrt{\frac{3}{2}} \) and \( x = \pm \sqrt{\frac{1}{2}} \). Simplify the square roots to get: \( x = \pm \frac{\sqrt{6}}{2} \) and \( x = \pm \frac{\sqrt{2}}{2} \)

Key Concepts

quadratic substitution

Quadratic substitution is a useful technique for solving higher-degree polynomial equations by transforming them into quadratic form. By introducing a new variable, we can simplify the equation, making it easier to solve. In this specific problem, we set a new variable, say, \(y\), such that \(y = x^2\). This reduces the original problem \(4x^4 - 8x^2 + 3 = 0\) into a more manageable quadratic equation: \(4y^2 - 8y + 3 = 0\). By recognizing patterns and making these substitutions, we can transform complicated expressions into forms that are straightforward to solve. This step is crucial for breaking down and systematically solving polynomial equations.

quadratic formula

The quadratic formula is a fundamental tool in algebra for solving quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is given by: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Here, \(a\), \(b\), and \(c\) are coefficients in the quadratic equation. For our problem, we identified \(a = 4\), \(b = -8\), and \(c = 3\). Plugging these into the quadratic formula allows us to find the solutions for \(y\). Specifically, we get two solutions: \( y = \frac{8 \pm \sqrt{16}}{8} = \frac{3}{2} \) and \( y = \frac{1}{2} \). The derivation involves computing the discriminant \(\Delta = b^2 - 4ac\) and then using it to find the roots. This process is not only mechanical but also highlights the elegance of algebraic methods in handling complex equations.

back-substitution

Back-substitution is the final step where we revert to our original variable after solving the quadratic equation. Once we find the solutions for our substituted variable, \(y\), we replace it back to express the solutions in terms of the original variable, \(x\). In this case, we solved for \(y\) and got \(y = \frac{3}{2}\) and \(y = \frac{1}{2}\). Back-substituting \(y = x^2\) gives us \(x^2 = \frac{3}{2}\) and \(x^2 = \frac{1}{2}\). Taking the square root of both sides, we obtain \(x = \pm \frac{\sqrt{6}}{2}\) and \(x = \pm \frac{\sqrt{2}}{2}\). This step ensures that we correctly interpret the results in the context of the original problem, thus completing the solution.