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Kaylee solves the equation below by first squaring both sides of the equation. 1-y=sqrt(2y^(2)-7) What extraneous solution does Kaylee obtain? y=

Kaylee solves the equation below by first squaring both sides of the equation.\newline1−y=2y2−7 1-y=\sqrt{2 y^{2}-7} \newlineWhat extraneous solution does Kaylee obtain?\newliney= y=

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Q. Kaylee solves the equation below by first squaring both sides of the equation.\newline1−y=2y2−7 1-y=\sqrt{2 y^{2}-7} \newlineWhat extraneous solution does Kaylee obtain?\newliney= y=
  1. Square both sides: Square both sides of the equation to eliminate the square root.\newline1−y=2y2−71 - y = \sqrt{2y^2 - 7}\newline(1−y)2=(2y2−7)2(1 - y)^2 = (\sqrt{2y^2 - 7})^2
  2. Simplify after squaring: Simplify both sides of the equation after squaring.\newline(1−y)2=2y2−7(1 - y)^2 = 2y^2 - 7\newline1−2y+y2=2y2−71 - 2y + y^2 = 2y^2 - 7
  3. Rearrange and set to zero: Rearrange the equation to bring all terms to one side and set the equation to zero.\newline0=2y2−y2−2y+7−10 = 2y^2 - y^2 - 2y + 7 - 1\newline0=y2−2y+60 = y^2 - 2y + 6
  4. Factor or use quadratic formula: Factor the quadratic equation, if possible, or use the quadratic formula to find the values of yy. However, the quadratic equation y2−2y+6y^2 - 2y + 6 does not factor nicely, so we will use the quadratic formula. y=−(−2)±(−2)2−4(1)(6)2(1)y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(6)}}{2(1)}
  5. Calculate discriminant: Calculate the discriminant to determine if there are real solutions.\newlineDiscriminant = (−2)2−4(1)(6)=4−24=−20(-2)^2 - 4(1)(6) = 4 - 24 = -20\newlineSince the discriminant is negative, there are no real solutions to the equation y2−2y+6=0y^2 - 2y + 6 = 0.
  6. Check for extraneous solution: Realize that the error occurred because the original equation did not have real solutions, but squaring both sides introduced an extraneous solution. We need to check the original equation with the solutions obtained from the quadratic equation to see which one is extraneous.
  7. No real solutions: Since the discriminant is negative, we made a mistake in our assumption that squaring both sides would yield a valid solution. There are no real solutions to the quadratic equation, so there cannot be an extraneous solution.

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