Problem 85 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$6(x+2)^{4}-11(x+2)^{2}=-4$$

Step-by-Step Solution

Answer

The solutions are \( x = -2 \pm \sqrt{\frac{4}{3}} \) and \( x = -2 \pm \sqrt{\frac{1}{2}} \).

Step 1: Introduce a substitution

Let \( y = (x + 2)^{2} \). This substitution will simplify the equation. Now, rewrite the given equation using \( y \).

Step 2: Rewrite the equation

Substituting \( y \) into the equation, we get: \( 6y^{2} - 11y = -4 \).

Step 3: Move all terms to one side

Add 4 to both sides to set the equation to zero: \( 6y^{2} - 11y + 4 = 0 \).

Step 4: Factor the quadratic equation

Factorize \( 6y^{2} - 11y + 4 \) into two binomials: \( (3y-4)(2y-1) = 0 \).

Step 5: Solve for \( y \)

Set each factor equal to zero and solve for \( y \):

Step 6: Solve \( 3y-4 = 0 \)

Solving \( 3y-4 = 0 \) yields \( y = \frac{4}{3} \).

Step 7: Solve \( 2y-1 = 0 \)

Solving \( 2y-1 = 0 \) yields \( y = \frac{1}{2} \).

Step 8: Substitute back \( y = (x+2)^{2} \)

Now, substitute back \( y \) into the original substitution. So, \( (x+2)^{2} = \frac{4}{3} \) or \( (x+2)^{2} = \frac{1}{2} \).

Step 9: Solve for \( x \) from \( (x+2)^{2} = \frac{4}{3} \)

Taking the square root of both sides, we get \( x + 2 = \pm \sqrt{\frac{4}{3}} \). This simplifies to two solutions: \( x = -2 \pm \sqrt{\frac{4}{3}} \).

Step 10: Solve for \( x \) from \( (x+2)^{2} = \frac{1}{2} \)

Taking the square root of both sides, we get \( x + 2 = \pm \sqrt{\frac{1}{2}} \). This simplifies to two solutions: \( x = -2 \pm \sqrt{\frac{1}{2}} \).

Step 11: Write the final solutions

The final solutions for the given equation are: \( x = -2 + \sqrt{\frac{4}{3}} \), \( x = -2 - \sqrt{\frac{4}{3}} \), \( x = -2 + \sqrt{\frac{1}{2}} \), and \( x = -2 - \sqrt{\frac{1}{2}} \).

Key Concepts

Substitution Method

The substitution method is a powerful technique to simplify complicated equations. By introducing a new variable, we can turn a complex problem into a simpler one. For instance, in the problem, we let \( y = (x + 2)^{2} \). This changes the original equation into a quadratic form: \( 6y^{2} - 11y = -4 \). The new equation is easier to handle because it is a standard quadratic. This method is especially useful when dealing with polynomial or rational equations. By replacing parts of the equation with a new variable, we simplify the original expression, making it easier to solve.

Factoring Quadratics

Factoring quadratics is a technique used to break down a quadratic equation into simpler, solvable parts. After substitution, the equation became \( 6y^{2} - 11y + 4 = 0 \). We then factored it into \( (3y-4)(2y-1) = 0 \). Factoring makes it easy to find the solutions: set each factor to zero and solve for \( y \). This process involves:

• Finding two numbers that multiply to the constant term (in this case, 4) and add to the coefficient of the middle term (in this case, -11).
• Using these numbers to break the middle term and factor by grouping.

Once the quadratic is factored into binomials, solving them individually gives the possible values for \( y \).

Completing the Square

Completing the square is a method to solve quadratic equations by transforming them into a perfect square trinomial. Although this method was not directly used in our problem, it's another way to solve quadratics. To complete the square:

• Ensure the quadratic equation is in the form \( ax^2 + bx + c = 0 \).
• Move the constant term to the other side: \( ax^2 + bx = -c \).
• If \( a \) is not 1, divide the entire equation by \( a \).
• Add and subtract \( \left( \frac{b}{2a} \right)^2 \) inside the equation.
• Rewrite the left side as a square of a binomial.

This method transforms the quadratic into an equation of the form \( (x+p)^2 = q \), which can be easily solved by taking the square root.

Radical Expressions

Radical expressions involve numbers and variables under the radical sign. In our solved problem, after substituting back, we ended up with \( (x + 2)^{2} = \frac{4}{3} \) and \( (x + 2)^{2} = \frac{1}{2} \). Solving these required taking the square root of both sides:

• \( x + 2 = \pm \sqrt{\frac{4}{3}} \)
• \( x + 2 = \pm \sqrt{\frac{1}{2}} \)

This step introduced radical expressions. Taking the square root of a fraction involves taking the square root of both the numerator and the denominator individually. Simplifying these square roots gave us the final solutions. Understanding how to work with radicals is crucial as they often appear in algebra and calculus.