Question

question 10 (3 points) solve the equation by completing the square (put answers in order from least to greatest and keep answers in fraction form): 9x² + 12x + 5 = 10 the square is (fx + g)² = h f = g = h = which means that x = __ or x = __ blank 1: blank 2: blank 3: blank 4: blank 5:

Explanation:

Step1: Divide the equation by 9

$$\begin{align*} 9x^{2}+12x + 5&=10\\ x^{2}+\frac{12}{9}x+\frac{5}{9}&=\frac{10}{9}\\ x^{2}+\frac{4}{3}x+\frac{5}{9}&=\frac{10}{9} \end{align*}$$

Step2: Move the constant term to the right - hand side

$$ x^{2}+\frac{4}{3}x=\frac{10}{9}-\frac{5}{9}=\frac{5}{9} $$

Step3: Complete the square

The coefficient of \(x\) is \(\frac{4}{3}\). Half of it is \(\frac{2}{3}\), and \((\frac{2}{3})^{2}=\frac{4}{9}\)

$$\begin{align*} x^{2}+\frac{4}{3}x+\frac{4}{9}&=\frac{5}{9}+\frac{4}{9}\\ (x + \frac{2}{3})^{2}&=1 \end{align*}$$

Step4: Solve for \(x\)

$$\begin{align*} x+\frac{2}{3}&=\pm1\\ x&=- \frac{2}{3}\pm1 \end{align*}$$

When \(x=-\frac{2}{3}+1=\frac{1}{3}\), when \(x=-\frac{2}{3}-1=-\frac{5}{3}\)

Answer:

Blank 1: \(-\frac{5}{3}\), \(\frac{1}{3}\)
Blank 2: \(1\)
Blank 3: \(\frac{2}{3}\)
Blank 4: \(\frac{4}{3}\)
Blank 5: \(\frac{2}{3}\)