Problem 57 - College Algebra

Question

Solve each equation in by making an appropriate substitution. $$ \left(y-\frac{8}{y}\right)^{2}+5\left(y-\frac{8}{y}\right)-14=0 $$

Step-by-Step Solution

Answer

Solving the equation yields four solutions over the reals: \(y=4-\sqrt{20}\), \(y=4+\sqrt{20}\), \(y=-8\), and \(y=1\).

Step 1: Substitution

Let us substitute \(z = y-\frac{8}{y}\). This change of variable transforms our equation into a standard quadratic equation of the form \(az^2+bz+c=0\): the equation becomes \(z^2+5z-14=0\).

Step 2: Solve the Quadratic Equation

This quadratic equation can be factored to \((z-2)(z+7)=0\). This gives two possible solutions for \(z\): \(z=2\) and \(z=-7\).

Step 3: Solving for y

Now we can substitute \(z\) back in terms of \(y\) for each solution of \(z\), and solve each equation. After simplifying the equation considering \(z = 2\), we obtain \(y^2-2y-8=0\), which can be solved to get \(y=4-\sqrt{20}\) and \(y=4+\sqrt{20}\). For \(z = -7\), the equation simplifies to \(y^2+7y-8=0\), the solutions of which are \(y=-8\) and \(y=1\).

Key Concepts

Algebraic Substitution

Algebraic substitution is a helpful technique, especially in solving complex equations. It involves replacing a part of an equation with a new variable to simplify the equation. In our original exercise, the expression \( y - \frac{8}{y} \) was substituted with \( z \). This transformed a complicated equation into a more standard quadratic form.

The primary goal of substitution is to reduce complexity and to handle components of the equation separately. By focusing on one part, it becomes easier to identify and apply solutions that might not have been immediately apparent in the original form.

In this example, substituting \( z = y - \frac{8}{y} \) gave us the equation \( z^2 + 5z - 14 = 0 \). It's crucial in substitution to rearrange and solve what you can to make later steps more straightforward.

Factoring

Factoring is another essential concept used to solve quadratic equations. It involves expressing a quadratic equation as a product of its linear factors. This can often uncover solutions more efficiently than traditional methods.

In our equation \( z^2 + 5z - 14 = 0 \), we factor it into \((z - 2)(z + 7) = 0\). This shows us that \( z \) could be either 2 or -7. Each factor represents a potential solution.

Factoring requires identifying two numbers that multiply to give the constant term while adding up to the coefficient of the middle term. This method is especially useful when the quadratic equation can be quickly and neatly split into linear terms.

Complex Numbers

In some quadratic equations, the solutions can involve complex numbers. However, in the exercise provided, our solutions were real numbers. But it's still important to understand what complex numbers are as they frequently arise in algebra.

Complex numbers consist of a real part and an imaginary part and are usually written in the form \( a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit, defined as \( \sqrt{-1} \).

Even though complex numbers were not part of our direct solutions, being prepared to handle them is an advantage, especially when real solutions do not exist for a particular quadratic equation.

Solving Polynomials

Solving polynomials, especially quadratics, is a foundational skill in algebra. It combines techniques like substitution, factoring, and sometimes completing the square or using the quadratic formula.

In the given exercise, solving for \( z \) with the equation \( z^2 + 5z - 14 = 0 \) allowed us to find values of \( z \) that led back to solving a polynomial equation in terms of \( y \). Using the solutions \( z = 2 \) and \( z = -7 \), we replaced \( z \) back into the original substitution \( y - \frac{8}{y} = z \), to find polynomial equations. Solving these for \( y \) revealed our final answers.

Polynomial solutions often require checking if substitutions and factors bring you back to the original values or equivalent forms. This confirms the correctness of the solution achieved by substituting not just once, but sometimes multiple times.