Problem 52 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \(x^{4}-37 x^{2}+36=0\)

Step-by-Step Solution

Answer

The solutions are \( x = 6 \), \( x = -6 \), \( x = 1 \), and \( x = -1 \).

Step 1: Substitute a new variable

Letting us simplify the equation by using substitution. Set \( y = x^2 \). This changes the original equation from \( x^4 - 37x^2 + 36 = 0 \) to \( y^2 - 37y + 36 = 0 \).

Step 2: Solve the quadratic equation

Solve the quadratic equation \( y^2 - 37y + 36 = 0 \) using the quadratic formula: \( y = \frac{-b \,\pm\, \sqrt{b^2 - 4ac}}{2a} \), where \(a = 1\), \(b = -37\), and \(c = 36\).

Step 3: Calculate the discriminant

Calculate the discriminant \( \Delta = b^2 - 4ac \): \( \Delta = (-37)^2 - 4(1)(36) = 1369 - 144 = 1225 \).

Step 4: Find the roots of the quadratic equation

Plugging the discriminant result back into the quadratic formula gives: \( y = \frac{37 \,\pm\, \sqrt{1225}}{2} = \frac{37 \,\pm\, 35}{2} \). This results in two solutions: \( y = 36 \) and \( y = 1 \).

Step 5: Reverse the substitution

Return to the original variable: if \( y = x^2 \), then solving \( x^2 = 36 \) gives \( x = \pm6 \); solving \( x^2 = 1 \) gives \( x = \pm1 \).

Step 6: Verify the solutions

Check all four solutions \( x = 6, -6, 1, -1 \) by plugging them into the original equation: \( x^4 - 37x^2 + 36 = 0 \). Each should satisfy the equation: For \( x = 6 \): \( 6^4 - 37 \cdot 6^2 + 36 = 0 \), true. For \( x = -6 \): same as \( x = 6 \), true. For \( x = 1 \): \( 1^4 - 37 \cdot 1^2 + 36 = 0 \), true. For \( x = -1 \): same as \( x = 1 \), true.

Key Concepts

Quadratic Substitution

Quadratic substitution is a clever technique to simplify complex polynomial equations. It involves substituting a variable to transform a higher-degree equation into a quadratic form. Let's look at our example: the equation is \(x^4 - 37x^2 + 36 = 0\). Here, you can let \(y = x^2\). This substitution turns our original equation into \(y^2 - 37y + 36 = 0\), a quadratic equation. This makes it much easier to solve. Once you find the roots for \(y\), you then substitute back to solve for \(x\). This process breaks a complicated problem into simpler steps.

Quadratic Formula

The quadratic formula is a powerful tool for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). The formula is as follows:
\(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).
Let's apply this to our substituted equation \(y^2 - 37y + 36 = 0\). Here, \(a = 1\), \(b = -37\), and \(c = 36\). After substituting these into the formula, we get:
\(y = \frac{37 \pm \sqrt{1225}}{2}\).
Solving this, we find two potential values for \(y\): \(y = 36\) and \(y = 1\). These solutions hold the key to finding the original values of \(x\).

Discriminant

The discriminant in the quadratic formula \(\Delta\) is given by the expression \(b^2 - 4ac\). It tells us about the nature of the roots of the quadratic equation. In our example, we calculate the discriminant as follows:
\(\Delta = (-37)^2 - 4(1)(36) = 1369 - 144 = 1225\).
Since the discriminant \(\Delta\) is positive and a perfect square, it indicates that the quadratic equation has two distinct real roots.
Understanding the discriminant helps us gauge the possible solutions even before using the quadratic formula. It's like a preview of the roots without fully solving the equation.