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

We want to solve the following equation.\newline∣x−4∣=x2−6x+9 |x-4|=x^{2}-6 x+9 \newlineOne of the solutions is x≈1.4 x \approx 1.4 .\newlineFind the other solution.\newlineHint: Use a graphing calculator.\newlineRound your answer to the nearest tenth.\newlinex≈ x \approx

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Q. We want to solve the following equation.\newline∣x−4∣=x2−6x+9 |x-4|=x^{2}-6 x+9 \newlineOne of the solutions is x≈1.4 x \approx 1.4 .\newlineFind the other solution.\newlineHint: Use a graphing calculator.\newlineRound your answer to the nearest tenth.\newlinex≈ x \approx
  1. Understanding the Equation: First, let's understand the equation ∣x−4∣=x2−6x+9|x-4| = x^2 - 6x + 9. The absolute value function ∣x−4∣|x-4| will split the equation into two cases, one for x≥4x \geq 4 and one for x < 4.
  2. Solving for x≥4x \geq 4: For x≥4x \geq 4, the equation becomes x−4=x2−6x+9x - 4 = x^2 - 6x + 9. Let's solve this quadratic equation.\newlinex−4=x2−6x+9x - 4 = x^2 - 6x + 9\newline0=x2−7x+130 = x^2 - 7x + 13\newlineThis is a quadratic equation in standard form.
  3. Using the Quadratic Formula: We can use the quadratic formula to solve for xx: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=1a = 1, b=−7b = -7, and c=13c = 13.
  4. No Real Solutions for x≥4x \geq 4: Plugging in the values, we get:\newlinex=7±49−4(1)(13)2(1)x = \frac{7 \pm \sqrt{49 - 4(1)(13)}}{2(1)}\newlinex=7±49−522x = \frac{7 \pm \sqrt{49 - 52}}{2}\newlinex=7±−32x = \frac{7 \pm \sqrt{-3}}{2}\newlineSince we cannot take the square root of a negative number in the real number system, there are no real solutions for x≥4x \geq 4.
  5. Solving for x < 4: For x < 4, the equation becomes −(x−4)=x2−6x+9- (x - 4) = x^2 - 6x + 9. Let's solve this quadratic equation.\newline−(x−4)=x2−6x+9- (x - 4) = x^2 - 6x + 9\newline−x+4=x2−6x+9- x + 4 = x^2 - 6x + 9\newline0=x2−5x+50 = x^2 - 5x + 5\newlineThis is another quadratic equation in standard form.
  6. Using the Quadratic Formula Again: Again, we use the quadratic formula to solve for xx: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=1a = 1, b=−5b = -5, and c=5c = 5.
  7. Two Potential Solutions: Plugging in the values, we get:\newlinex=5±25−4(1)(5)2(1)x = \frac{5 \pm \sqrt{25 - 4(1)(5)}}{2(1)}\newlinex=5±25−202x = \frac{5 \pm \sqrt{25 - 20}}{2}\newlinex=5±52x = \frac{5 \pm \sqrt{5}}{2}\newlineWe have two potential solutions here, but we must remember that we are looking for the solution where x < 4.
  8. Estimating the Solutions: The two potential solutions are: \newlinex=5+52x = \frac{5 + \sqrt{5}}{2} and x=5−52x = \frac{5 - \sqrt{5}}{2}\newlineSince 5\sqrt{5} is approximately 2.22.2, we can estimate these solutions.\newlinex=5+2.22≈3.6x = \frac{5 + 2.2}{2} \approx 3.6\newlinex=5−2.22≈1.4x = \frac{5 - 2.2}{2} \approx 1.4\newlineWe were given that x≈1.4x \approx 1.4 is one solution, so the other solution must be x≈3.6x \approx 3.6.
  9. Verifying the Solution: However, we need to verify that this solution is indeed less than 44. Since 3.63.6 is less than 44, it satisfies the condition for the second case (x < 4).

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