We want to solve the following equation. $\newline$ $|x-4|=x^{2}-6 x+9$ $\newline$ One of the solutions is $x \approx 1.4$ . $\newline$ Find the other solution. $\newline$ Hint: Use a graphing calculator. $\newline$ Round your answer to the nearest tenth. $\newline$ $x \approx$

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Algebra 1

Transformations of absolute value functions: translations and reflections

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Q. We want to solve the following equation.

\newline

|x-4|=x^{2}-6 x+9

\newline

One of the solutions is

x \approx 1.4

\newline

Find the other solution.

\newline

Hint: Use a graphing calculator.

\newline

Round your answer to the nearest tenth.

\newline

x \approx

Understanding the Equation: First, let's understand the equation $|x-4| = x^2 - 6x + 9$ . The absolute value function $|x-4|$ will split the equation into two cases, one for $x \geq 4$ and one for $x < 4$ .
Solving for $x \geq 4$ : For $x \geq 4$ , the equation becomes $x - 4 = x^2 - 6x + 9$ . Let's solve this quadratic equation. $\newline$ $x - 4 = x^2 - 6x + 9$ $\newline$ $0 = x^2 - 7x + 13$ $\newline$ This is a quadratic equation in standard form.
Using the Quadratic Formula: We can use the quadratic formula to solve for $x$ : $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = -7$ , and $c = 13$ .
No Real Solutions for $x \geq 4$ : Plugging in the values, we get: $\newline$ $x = \frac{7 \pm \sqrt{49 - 4(1)(13)}}{2(1)}$ $\newline$ $x = \frac{7 \pm \sqrt{49 - 52}}{2}$ $\newline$ $x = \frac{7 \pm \sqrt{-3}}{2}$ $\newline$ Since we cannot take the square root of a negative number in the real number system, there are no real solutions for $x \geq 4$ .
Solving for $x < 4$ : For $x < 4$ , the equation becomes $- (x - 4) = x^2 - 6x + 9$ . Let's solve this quadratic equation. $\newline$ $- (x - 4) = x^2 - 6x + 9$ $\newline$ $- x + 4 = x^2 - 6x + 9$ $\newline$ $0 = x^2 - 5x + 5$ $\newline$ This is another quadratic equation in standard form.
Using the Quadratic Formula Again: Again, we use the quadratic formula to solve for $x$ : $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = -5$ , and $c = 5$ .
Two Potential Solutions: Plugging in the values, we get: $\newline$ $x = \frac{5 \pm \sqrt{25 - 4(1)(5)}}{2(1)}$ $\newline$ $x = \frac{5 \pm \sqrt{25 - 20}}{2}$ $\newline$ $x = \frac{5 \pm \sqrt{5}}{2}$ $\newline$ We have two potential solutions here, but we must remember that we are looking for the solution where $x < 4$ .
Estimating the Solutions: The two potential solutions are: $\newline$ $x = \frac{5 + \sqrt{5}}{2}$ and $x = \frac{5 - \sqrt{5}}{2}$ $\newline$ Since $\sqrt{5}$ is approximately $2.2$ , we can estimate these solutions. $\newline$ $x = \frac{5 + 2.2}{2} \approx 3.6$ $\newline$ $x = \frac{5 - 2.2}{2} \approx 1.4$ $\newline$ We were given that $x \approx 1.4$ is one solution, so the other solution must be $x \approx 3.6$ .
Verifying the Solution: However, we need to verify that this solution is indeed less than $4$ . Since $3.6$ is less than $4$ , it satisfies the condition for the second case ( $x < 4$ ).