Problem 39 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the equation by using the quadratic formula. $$ y^{4}-5 y^{2}+6=0 $$

Step-by-Step Solution

Answer

The possible values of y are \(y = \sqrt{3}, -\sqrt{3}, \sqrt{2}, -\sqrt{2}\), after solving the quadratic equation in disguise using the quadratic formula.

Step 1: Identify the quadratic equation

Rewrite the given quartic equation as a quadratic equation in terms of \(z = y^2\): \[z^2 - 5z + 6 = 0\]

Step 2: Apply the quadratic formula to find z

Apply the quadratic formula, which is given by \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -5\), and \(c = 6\): \[z = \frac{-(-5) \pm \sqrt{(-5)^2 - (4)(1)(6)}}{2(1)}\]

Step 3: Calculate discriminant (Δ)

Calculate the discriminant, which is given by \(\Delta = b^2 - 4ac\): \[\Delta = (-5)^2 - (4)(1)(6) = 25 - 24 = 1\]

Step 4: Determine the values of z

Replace the discriminant in the quadratic formula with the calculated value (1) to find the values of z: \[z = \frac{5 \pm \sqrt{1}}{2}\] We find two possible values for z: \[z_1 = \frac{5 + \sqrt{1}}{2} = \frac{5 + 1}{2} = 3\] and \[z_2 = \frac{5 - \sqrt{1}}{2} = \frac{5 - 1}{2} = 2\]

Step 5: Find the possible values of y

Because we have two possible values for z, we need to find the corresponding values of y according to each z value. Remember that \(y^2 = z\), so we will solve \(y^2 = z_1\) and \(y^2 = z_2\) and find the square root of each z value: For z = 3, the possible values of y are: \[y = \pm\sqrt{3}\] For z = 2, the possible values of y are: \[y = \pm\sqrt{2}\] So, the possible values of y are: \[y = \sqrt{3}, -\sqrt{3}, \sqrt{2}, -\sqrt{2}\]

Key Concepts

Understanding Quartic Equations

Quartic equations are equations of the fourth degree, which means their highest power of the variable is four. In the original math problem, the quartic equation is given as: \[y^4 - 5y^2 + 6 = 0\]To solve a quartic equation, one approach is to use a substitution to transform it into a simpler quadratic form. By letting \(z = y^2\), the equation simplifies to:\[z^2 - 5z + 6 = 0\]This transformation effectively turns the quartic equation into a quadratic one, making it easier to solve using the quadratic formula. When faced with a quartic equation, looking for patterns or substitutions that can reduce the complexity of the problem is often a successful strategy.By reducing the quartic to a quadratic, we allow ourselves to use a well-known formula to find the solutions with ease.

Calculating the Discriminant

The discriminant is a key part of the quadratic formula, and it helps determine the nature of the roots of a quadratic equation. For any quadratic equation of the form \(ax^2 + bx + c = 0\), the discriminant \(\Delta\) is calculated as:\[\Delta = b^2 - 4ac\]In our example, the quadratic equation \(z^2 - 5z + 6 = 0\) has:

• \(a = 1\)
• \(b = -5\)
• \(c = 6\)

By substituting these into the discriminant formula, we get:\[\Delta = (-5)^2 - 4 \times 1 \times 6 = 25 - 24 = 1\]A positive discriminant, like 1 in this example, means that the quadratic equation has two distinct real roots. This provides us with crucial information before proceeding to solve the equation.

Solutions of Quadratic Equations

Once you have the discriminant, solving a quadratic equation using the quadratic formula is straightforward. The formula is: \[z = \frac{-b \pm \sqrt{\Delta}}{2a}\]In the context of our problem, with \(a = 1\), \(b = -5\), and \(\Delta = 1\), we find the values of \(z\) by substituting these values into the formula:\[z = \frac{5 \pm \sqrt{1}}{2}\]This results in two solutions:

• \(z_1 = \frac{5 + 1}{2} = 3\)
• \(z_2 = \frac{5 - 1}{2} = 2\)

To find the solutions for \(y\), remember \(z = y^2\). Therefore, for each \(z\), we solve \(y^2 = z\):

• For \(z = 3\), \(y = \pm \sqrt{3}\)
• For \(z = 2\), \(y = \pm \sqrt{2}\)

Hence, the solutions of the original quartic equation are:\[y = \sqrt{3}, -\sqrt{3}, \sqrt{2}, -\sqrt{2}\]Using the quadratic formula in conjunction with the understanding of the discriminant makes solving even complex equations like quartics feasible and systematic.