Problem 43 - Precalculus Student Solutions Manual 5th

Question

Solve each equation by completing the square. $$2 x^{2}+x=10$$

Step-by-Step Solution

Answer

x = 2 or x = -\frac{5}{2}

Step 1: Move the constant term to the right side

Rewrite the equation so that the constant term is on the right side of the equation. This gives: \[2x^{2}+x=10\] Then, subtract 10 from both sides: \[2x^{2} + x - 10 = 0\] Add 10 to both sides to isolate the variable terms: \[2x^{2} + x = 10\]

Step 2: Divide by the coefficient of \(x^2\)

Divide every term in the equation by 2 to make the coefficient of \(x^2\) equal to 1: \[\frac{2x^{2}}{2} + \frac{x}{2} = \frac{10}{2}\] which simplifies to: \[ x^{2} + \frac{1}{2} x = 5\]

Step 3: Move the constant term to the right side again

Subtract 5 from both sides to isolate the terms involving x: \[ x^{2} + \frac{1}{2} x = 5 \] Now, subtract 5: \[ x^{2} + \frac{1}{2} x - 5 = 0\] add 5 to both sides: \[ x^{2} + \frac{1}{2} x = 5\]

Step 4: Complete the square

To complete the square, take half of the coefficient of x, square it, and add to both sides. The coefficient of x is \(\frac{1}{2}\), half of it is \(\frac{1}{4}\), and squaring \(\frac{1}{4}\) results in \(\frac{1}{16}\). Therefore, add \(\frac{1}{16}\) to both sides: \[ x^{2} + \frac{1}{2} x + \frac{1}{16} = 5 + \frac{1}{16}\] \[ x^{2} + \frac{1}{2} x + \frac{1}{16} = \frac{81}{16}\]

Step 5: Factor the left side and simplify the right side

Factor the left side as a perfect square trinomial and simplify the right side: \[ \left( x + \frac{1}{4}\right)^{2} = \frac{81}{16}\]

Step 6: Solve the equation

Take the square root of both sides of the equation: \[ x + \frac{1}{4} = \pm\sqrt{ \frac{81}{16}}\] Simplify the square root: \[ x + \frac{1}{4} = \pm \frac{9}{4}\]

Step 7: Solve for x

Solve for x by subtracting \(\frac{1}{4}\) from both sides: \[ x = -\frac{1}{4} + \frac{9}{4} \quad \text{or} \quad x = -\frac{1}{4} - \frac{9}{4} \] which results in: \[ x = 2 \quad \text{or} \quad x = -\frac{5}{2} \]

Key Concepts

quadratic equations

Quadratic equations take the form \[ ax^2 + bx + c = 0 \]. These equations feature an \(x^2\) term (the quadratic term), an \(x\) term (the linear term), and a constant term \(c\). The main goal when solving quadratic equations is to find the values of \(x\) that make the equation true.

Quadratic equations are a key part of algebra because they appear in various contexts like projectile motion in physics and area problems in geometry.

• The standard form: \(ax^2 + bx + c = 0\)
• The general solution involves finding the roots

To solve these equations, you can use several methods: factoring, the quadratic formula, and completing the square. Each method has its own advantages depending on the specific equation you are dealing with. Completing the square is particularly useful when factoring is difficult or when you want to derive the quadratic formula yourself.

factoring

Factoring is a technique to solve quadratic equations by writing the quadratic expression as a product of two binomials. For some quadratic equations, this method is fast and efficient.

Let's consider the equation \[ x^2 + 5x + 6 = 0 \]. You would look for two numbers that multiply to 6 (the constant term) and add to 5 (the coefficient of the \(x\) term). Here, the numbers 2 and 3 work: \[ (x+2)(x+3) = 0 \].

You then set each factor to zero and solve for \(x\): \[ x+2=0 \Rightarrow x=-2 \] \[ x+3=0 \Rightarrow x=-3 \].

• Step 1: Look for two numbers that multiply to \(ac\) and add to \(b\)
• Step 2: Write the equation as a product of binomials
• Step 3: Set each factor equal to zero
• Step 4: Solve for \(x\)

However, not all quadratics can be easily factored, at which point methods like completing the square or using the quadratic formula become useful.

solving equations

Solving equations means finding all values that make the equation true. For quadratic equations, it involves finding the \(x\) values where the expression equals zero.

One powerful technique to solve equations is completing the square, which involves transforming a quadratic equation into a perfect square trinomial.

• First, ensure the coefficient of \(x^2\) is 1 by dividing through by \(a\)
• Move the constant term to the other side
• Add and subtract the square of half the \(x\) coefficient
• Factor the perfect square trinomial
• Take the square root of both sides
• Solve for \(x\)

For example, given \[ 2x^2 + x = 10 \], we follow the steps to get \[ \begin{aligned} & x^2 + \frac{1}{2}x = 5 \ \to & x^2 + \frac{1}{2}x + \frac{1}{16} = 5 + \frac{1}{16} \ \to & \big(x + \frac{1}{4}\big)^2 = \frac{81}{16} \ \to & x + \frac{1}{4} = \frac{9}{4} \ \to & x = 2 \text{ or } x = -\frac{5}{2} \ \text{.} \ \text{.} \ \text{.} \ \text{.}\]This method always works and is excellent for quadratic equations that aren't easily factored.