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Addison solves the equation below by first squaring both sides of the equation. 2x-1=sqrt(8-x) What extraneous solution does Addison obtain? x=

Addison solves the equation below by first squaring both sides of the equation.\newline2x−1=8−x 2 x-1=\sqrt{8-x} \newlineWhat extraneous solution does Addison obtain?\newlinex= x=

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Q. Addison solves the equation below by first squaring both sides of the equation.\newline2x−1=8−x 2 x-1=\sqrt{8-x} \newlineWhat extraneous solution does Addison obtain?\newlinex= x=
  1. Write Equation: Write down the original equation.\newline2x−1=8−x2x - 1 = \sqrt{8 - x}
  2. Square Both Sides: Square both sides of the equation to eliminate the square root.\newline(2x−1)2=(8−x)2(2x - 1)^2 = (\sqrt{8 - x})^2
  3. Perform Squaring: Perform the squaring on both sides.\newline(2x−1)(2x−1)=8−x(2x - 1)(2x - 1) = 8 - x
  4. Expand Left Side: Expand the left side of the equation using the FOIL method (First, Outer, Inner, Last).\newline4x2−2x−2x+1=8−x4x^2 - 2x - 2x + 1 = 8 - x
  5. Combine Like Terms: Combine like terms on the left side of the equation.\newline4x2−4x+1=8−x4x^2 - 4x + 1 = 8 - x
  6. Add xx to Both Sides: Add xx to both sides of the equation to bring all xx terms to one side.\newline4x2−4x+x+1=84x^2 - 4x + x + 1 = 8
  7. Subtract 88: Combine like terms on the left side of the equation.4x2−3x+1=84x^2 - 3x + 1 = 8
  8. Factor or Use Quadratic Formula: Subtract 88 from both sides of the equation to set the equation to zero.\newline4x2−3x+1−8=04x^2 - 3x + 1 - 8 = 0
  9. Calculate Discriminant: Combine like terms on the left side of the equation. 4x2−3x−7=04x^2 - 3x - 7 = 0
  10. Find Solutions: Factor the quadratic equation, if possible, or use the quadratic formula to find the solutions for xx. The quadratic formula is x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=4a = 4, b=−3b = -3, and c=−7c = -7.
  11. Check Solutions: Calculate the discriminant (b2−4acb^2 - 4ac) to determine the nature of the roots.\newlineDiscriminant = (−3)2−4(4)(−7)(-3)^2 - 4(4)(-7)\newlineDiscriminant = 9+1129 + 112\newlineDiscriminant = 121121
  12. Extraneous Solution: Since the discriminant is positive, there are two real solutions. Calculate the solutions using the quadratic formula.\newlinex=3±1218x = \frac{3 \pm \sqrt{121}}{8}\newlinex=3±118x = \frac{3 \pm 11}{8}
  13. Extraneous Solution: Since the discriminant is positive, there are two real solutions. Calculate the solutions using the quadratic formula.\newlinex=3±1218x = \frac{3 \pm \sqrt{121}}{8}\newlinex=3±118x = \frac{3 \pm 11}{8}Find the two solutions.\newlinex=(3+11)8x = \frac{(3 + 11)}{8} or x=(3−11)8x = \frac{(3 - 11)}{8}\newlinex=148x = \frac{14}{8} or x=−88x = \frac{-8}{8}\newlinex=1.75x = 1.75 or x=−1x = -1
  14. Extraneous Solution: Since the discriminant is positive, there are two real solutions. Calculate the solutions using the quadratic formula.\newlinex=3±1218x = \frac{3 \pm \sqrt{121}}{8}\newlinex=3±118x = \frac{3 \pm 11}{8} Find the two solutions.\newlinex=(3+11)8x = \frac{(3 + 11)}{8} or x=(3−11)8x = \frac{(3 - 11)}{8}\newlinex=148x = \frac{14}{8} or x=−88x = \frac{-8}{8}\newlinex=1.75x = 1.75 or x=−1x = -1 Check both solutions in the original equation to see if any of them is extraneous.\newlineFor x=1.75x = 1.75:\newline2(1.75)−1=8−1.752(1.75) - 1 = \sqrt{8 - 1.75}\newlinex=3±118x = \frac{3 \pm 11}{8}00\newlinex=3±118x = \frac{3 \pm 11}{8}11 (True)\newlineFor x=−1x = -1:\newlinex=3±118x = \frac{3 \pm 11}{8}33\newlinex=3±118x = \frac{3 \pm 11}{8}44\newlinex=3±118x = \frac{3 \pm 11}{8}55 (False)
  15. Extraneous Solution: Since the discriminant is positive, there are two real solutions. Calculate the solutions using the quadratic formula.\newlinex=3±1218x = \frac{3 \pm \sqrt{121}}{8}\newlinex=3±118x = \frac{3 \pm 11}{8}Find the two solutions.\newlinex=3+118x = \frac{3 + 11}{8} or x=3−118x = \frac{3 - 11}{8}\newlinex=148x = \frac{14}{8} or x=−88x = \frac{-8}{8}\newlinex=1.75x = 1.75 or x=−1x = -1Check both solutions in the original equation to see if any of them is extraneous.\newlineFor x=1.75x = 1.75:\newline2(1.75)−1=8−1.752(1.75) - 1 = \sqrt{8 - 1.75}\newlinex=3±118x = \frac{3 \pm 11}{8}00\newlinex=3±118x = \frac{3 \pm 11}{8}11 (True)\newlineFor x=−1x = -1:\newlinex=3±118x = \frac{3 \pm 11}{8}33\newlinex=3±118x = \frac{3 \pm 11}{8}44\newlinex=3±118x = \frac{3 \pm 11}{8}55 (False)The solution x=−1x = -1 does not satisfy the original equation, so it is the extraneous solution.\newlinex=−1x = -1

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