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Antonio is solving the following equation for y. -1=sqrt(y+11)-y His first few steps are given below. {:[y-1=sqrt(y+11)],[(y-1)^(2)=(sqrt(y+11))^(2)],[y^(2)-2y+1=y+11]:} Is it necessary for Antonio to check his answers for extraneous solutions? Choose 1 answer: (A) Yes (B) No

Antonio is solving the following equation for y y .\newline−1=y+11−y -1=\sqrt{y+11}-y \newlineHis first few steps are given below.\newliney−1amp;=y+11(y−1)2amp;=(y+11)2y2−2y+1amp;=y+11 \begin{aligned} y-1 & =\sqrt{y+11} \\ (y-1)^{2} & =(\sqrt{y+11})^{2} \\ y^{2}-2 y+1 & =y+11 \end{aligned} \newlineIs it necessary for Antonio to check his answers for extraneous solutions?\newlineChoose 11 answer:\newline(A) Yes\newline(B) No

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Q. Antonio is solving the following equation for y y .\newline−1=y+11−y -1=\sqrt{y+11}-y \newlineHis first few steps are given below.\newliney−1=y+11(y−1)2=(y+11)2y2−2y+1=y+11 \begin{aligned} y-1 & =\sqrt{y+11} \\ (y-1)^{2} & =(\sqrt{y+11})^{2} \\ y^{2}-2 y+1 & =y+11 \end{aligned} \newlineIs it necessary for Antonio to check his answers for extraneous solutions?\newlineChoose 11 answer:\newline(A) Yes\newline(B) No
  1. Isolate square root term: Antonio starts with the equation −1=y+11−y-1 = \sqrt{y + 11} - y. To isolate the square root term, he adds yy to both sides of the equation.\newline−1+y=y+11-1 + y = \sqrt{y + 11}\newlineThis gives us y−1=y+11y - 1 = \sqrt{y + 11}.
  2. Eliminate square root: Next, Antonio squares both sides of the equation to eliminate the square root. This is a critical step because squaring both sides of an equation can introduce extraneous solutions.\newline(y−1)2=(y+11)2(y - 1)^2 = (\sqrt{y + 11})^2\newlineThis simplifies to y2−2y+1=y+11y^2 - 2y + 1 = y + 11.
  3. Set equation to zero: Antonio now has a quadratic equation y2−2y+1=y+11y^2 - 2y + 1 = y + 11. He should subtract yy and 1111 from both sides to set the equation to zero and solve for yy.\newliney2−2y+1−y−11=0y^2 - 2y + 1 - y - 11 = 0\newlineThis simplifies to y2−3y−10=0y^2 - 3y - 10 = 0.
  4. Check for extraneous solutions: At this point, Antonio would typically factor the quadratic equation or use the quadratic formula to find the values of yy. However, the question asks if it is necessary to check for extraneous solutions. Since Antonio squared both sides of the equation, which is an operation that can introduce extraneous solutions, the answer is yes, it is necessary for Antonio to check his answers for extraneous solutions.

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