Problem 10 - Precalculus Student Solutions Manual 5th

Question

Use Choices \(A-D\) to answer each question. A. \(3 x^{2}-17 x-6=0\) B. \((2 x+5)^{2}=7\) C. \(x^{2}+x=12\) D. \((3 x-1)(x-7)=0\) Which equation is set up for direct use of the square root property? Solve it.

Step-by-Step Solution

Answer

The equation is \( (2x + 5)^2 = 7 \. The solutions are \ x = \frac{\sqrt{7} - 5}{2} \ and \ x = \frac{-\sqrt{7} - 5}{2} \.

Step 1: Identify the Equation with the Square Root Property

Review options A through D to identify the equation that can be solved by isolating a perfect square on one side of the equation, suitable for applying the square root property. Choice B, \[(2x+5)^2 = 7\], can be solved directly using the square root property.

Step 2: Isolate the Square Root

Set up the chosen equation for applying the square root property by taking the square root of both sides: \[ \sqrt{(2x + 5)^2} = \sqrt{7} \].

Step 3: Simplify Both Sides of the Equation

Simplify both sides: \[ 2x + 5 = \pm \sqrt{7} \]. This gives two separate equations: \ 2x + 5 = \sqrt{7} \ and \ 2x + 5 = -\sqrt{7} \.

Step 4: Solve for x

Solve each equation separately: \[ 2x + 5 = \sqrt{7} \ Rightarrow 2x = \sqrt{7} - 5 \ Rightarrow x = \frac{\sqrt{7} - 5}{2} \] and \[ 2x + 5 = -\sqrt{7} \ Rightarrow 2x = -\sqrt{7} - 5 \ Rightarrow x = \frac{-\sqrt{7} - 5}{2} \].

Key Concepts

Quadratic Equations

Quadratic equations are second-degree polynomial equations in one variable. They take the form ax^2 + bx + c = 0\,where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. The key identifying feature of a quadratic equation is the \(x^2\) term.

Solving quadratic equations can be done in several ways:

• Factoring
• Completing the square
• Using the quadratic formula
• Taking square roots (square root property)

In this exercise, option B \((2x+5)^2 = 7\)is a perfect example where we use the square root property directly. It simplifies the equation-solving process when we have a square term isolated, ready for square rooting.

Solving Equations

Solving equations involves finding the value(s) of the variable that make the equation true. Let's break down solving the given equation \((2x+5)^2 = 7\)using the square root property:

• Step 1: Identify the structure. Noticing that the equation is in the form \((something)^2 = number\)allows us to use the square root property.
• Step 2: Isolate the squared term by taking the square root of both sides, giving us two cases:
  \(2x + 5 = \sqrt{7}\) and
  \(2x + 5 = -\sqrt{7}\)
• Step 3: Solve each resulting linear equation. For \(2x + 5 = \sqrt{7}\):
  • Subtract 5 from both sides: \(2x = \sqrt{7} - 5\)
  • Divide by 2: \(x = \frac{\sqrt{7} - 5}{2}\)
• For \(2x + 5 = -\sqrt{7}\):
  • Subtract 5 from both sides: \(2x = -\sqrt{7} - 5\)
  • Divide by 2: \(x = \frac{-\sqrt{7} - 5}{2}\)

This method efficiently finds the solutions by managing the equation with simpler steps.

Algebraic Manipulation

Algebraic manipulation is crucial for solving equations. It involves rearranging and simplifying expressions to isolate the variable. Let's revisit the steps for \(2x + 5 = \sqrt{7}\) :

• Subtract a constant term:
Starting from \(2x + 5 = \sqrt{7}\), subtracting 5 from both sides gives us \(2x = \sqrt{7} - 5\)
• Divide by a coefficient:
To isolate \(x\), divide both sides by 2: \(x = \frac{\sqrt{7} - 5}{2}\).
• Handle square roots carefully:
Remember, square roots yield positive and negative values. Thus, we consider both \(+\sqrt{7}\) and \(-\sqrt{7}\).

Good algebraic manipulation skills make problem-solving more straightforward and less prone to error. They help in isolating variables step-by-step, applying properties like the square root property seamlessly.