Miku solves the equation below by first squaring both sides of the equation. $\newline$ $\sqrt{z^{2}+2 z-3}=z-3$ $\newline$ What extraneous solution does Miku obtain? $\newline$ $z=$

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Q. Miku solves the equation below by first squaring both sides of the equation.

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\sqrt{z^{2}+2 z-3}=z-3

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What extraneous solution does Miku obtain?

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z=

Rephrasing the Problem: First, let's rephrase the ＂What extraneous solution does Miku obtain when solving the equation $\sqrt{z^{2}+2z-3}=z-3$ by first squaring both sides?＂
Squaring Both Sides: Miku starts by squaring both sides of the equation to eliminate the square root. The equation becomes $(\sqrt{z^{2}+2z-3})^2 = (z-3)^2$ .
Expanding the Equation: After squaring both sides, the equation simplifies to $z^{2}+2z-3 = (z-3)^{2}$ . We need to expand the right side of the equation.
Solving for z: Expanding the right side, we get $z^{2}+2z-3 = z^2 - 6z + 9$ .
Checking for Extraneous Solution: Next, we subtract $z^2$ from both sides to simplify the equation, which gives us $2z-3 = -6z + 9$ .
Checking for Extraneous Solution: Next, we subtract $z^2$ from both sides to simplify the equation, which gives us $2z-3 = -6z + 9$ .Now, we add $6z$ to both sides to isolate the $z$ terms on one side, resulting in $2z + 6z - 3 = 9$ .
Checking for Extraneous Solution: Next, we subtract $z^2$ from both sides to simplify the equation, which gives us $2z-3 = -6z + 9$ .Now, we add $6z$ to both sides to isolate the $z$ terms on one side, resulting in $2z + 6z - 3 = 9$ .Combining like terms, we get $8z - 3 = 9$ .
Checking for Extraneous Solution: Next, we subtract $z^2$ from both sides to simplify the equation, which gives us $2z-3 = -6z + 9$ .Now, we add $6z$ to both sides to isolate the $z$ terms on one side, resulting in $2z + 6z - 3 = 9$ .Combining like terms, we get $8z - 3 = 9$ .Adding $3$ to both sides gives us $8z = 12$ .
Checking for Extraneous Solution: Next, we subtract $z^2$ from both sides to simplify the equation, which gives us $2z-3 = -6z + 9$ .Now, we add $6z$ to both sides to isolate the $z$ terms on one side, resulting in $2z + 6z - 3 = 9$ .Combining like terms, we get $8z - 3 = 9$ .Adding $3$ to both sides gives us $8z = 12$ .Finally, we divide both sides by $8$ to solve for $z$ , which gives us $2z-3 = -6z + 9$ $0$ .
Checking for Extraneous Solution: Next, we subtract $z^2$ from both sides to simplify the equation, which gives us $2z-3 = -6z + 9$ .Now, we add $6z$ to both sides to isolate the $z$ terms on one side, resulting in $2z + 6z - 3 = 9$ .Combining like terms, we get $8z - 3 = 9$ .Adding $3$ to both sides gives us $8z = 12$ .Finally, we divide both sides by $8$ to solve for $z$ , which gives us $2z-3 = -6z + 9$ $0$ .Simplifying the fraction $2z-3 = -6z + 9$ $1$ , we get $2z-3 = -6z + 9$ $2$ or $2z-3 = -6z + 9$ $3$ . This is the solution we obtain after squaring both sides of the equation.
Checking for Extraneous Solution: Next, we subtract $z^2$ from both sides to simplify the equation, which gives us $2z-3 = -6z + 9$ .Now, we add $6z$ to both sides to isolate the $z$ terms on one side, resulting in $2z + 6z - 3 = 9$ .Combining like terms, we get $8z - 3 = 9$ .Adding $3$ to both sides gives us $8z = 12$ .Finally, we divide both sides by $8$ to solve for $z$ , which gives us $2z-3 = -6z + 9$ $0$ .Simplifying the fraction $2z-3 = -6z + 9$ $1$ , we get $2z-3 = -6z + 9$ $2$ or $2z-3 = -6z + 9$ $3$ . This is the solution we obtain after squaring both sides of the equation.However, we must check this solution in the original equation to ensure it is not extraneous. We substitute $2z-3 = -6z + 9$ $2$ into the original equation $2z-3 = -6z + 9$ $5$ .
Checking for Extraneous Solution: Next, we subtract $z^2$ from both sides to simplify the equation, which gives us $2z-3 = -6z + 9$ . Now, we add $6z$ to both sides to isolate the $z$ terms on one side, resulting in $2z + 6z - 3 = 9$ . Combining like terms, we get $8z - 3 = 9$ . Adding $3$ to both sides gives us $8z = 12$ . Finally, we divide both sides by $8$ to solve for $z$ , which gives us $2z-3 = -6z + 9$ $0$ . Simplifying the fraction $2z-3 = -6z + 9$ $1$ , we get $2z-3 = -6z + 9$ $2$ or $2z-3 = -6z + 9$ $3$ . This is the solution we obtain after squaring both sides of the equation. However, we must check this solution in the original equation to ensure it is not extraneous. We substitute $2z-3 = -6z + 9$ $2$ into the original equation $2z-3 = -6z + 9$ $5$ . Substituting $2z-3 = -6z + 9$ $2$ into the left side of the original equation, we get $2z-3 = -6z + 9$ $7$ .
Checking for Extraneous Solution: Next, we subtract $z^2$ from both sides to simplify the equation, which gives us $2z-3 = -6z + 9$ .Now, we add $6z$ to both sides to isolate the $z$ terms on one side, resulting in $2z + 6z - 3 = 9$ .Combining like terms, we get $8z - 3 = 9$ .Adding $3$ to both sides gives us $8z = 12$ .Finally, we divide both sides by $8$ to solve for $z$ , which gives us $2z-3 = -6z + 9$ $0$ .Simplifying the fraction $2z-3 = -6z + 9$ $1$ , we get $2z-3 = -6z + 9$ $2$ or $2z-3 = -6z + 9$ $3$ . This is the solution we obtain after squaring both sides of the equation.However, we must check this solution in the original equation to ensure it is not extraneous. We substitute $2z-3 = -6z + 9$ $2$ into the original equation $2z-3 = -6z + 9$ $5$ .Substituting $2z-3 = -6z + 9$ $2$ into the left side of the original equation, we get $2z-3 = -6z + 9$ $7$ .Calculating the inside of the square root, we have $2z-3 = -6z + 9$ $8$ .
Checking for Extraneous Solution: Next, we subtract $z^2$ from both sides to simplify the equation, which gives us $2z-3 = -6z + 9$ .Now, we add $6z$ to both sides to isolate the $z$ terms on one side, resulting in $2z + 6z - 3 = 9$ .Combining like terms, we get $8z - 3 = 9$ .Adding $3$ to both sides gives us $8z = 12$ .Finally, we divide both sides by $8$ to solve for $z$ , which gives us $2z-3 = -6z + 9$ $0$ .Simplifying the fraction $2z-3 = -6z + 9$ $1$ , we get $2z-3 = -6z + 9$ $2$ or $2z-3 = -6z + 9$ $3$ . This is the solution we obtain after squaring both sides of the equation.However, we must check this solution in the original equation to ensure it is not extraneous. We substitute $2z-3 = -6z + 9$ $2$ into the original equation $2z-3 = -6z + 9$ $5$ .Substituting $2z-3 = -6z + 9$ $2$ into the left side of the original equation, we get $2z-3 = -6z + 9$ $7$ .Calculating the inside of the square root, we have $2z-3 = -6z + 9$ $8$ .Simplifying the terms inside the square root, we get $2z-3 = -6z + 9$ $9$ .
Checking for Extraneous Solution: Next, we subtract $z^2$ from both sides to simplify the equation, which gives us $2z-3 = -6z + 9$ . Now, we add $6z$ to both sides to isolate the $z$ terms on one side, resulting in $2z + 6z - 3 = 9$ . Combining like terms, we get $8z - 3 = 9$ . Adding $3$ to both sides gives us $8z = 12$ . Finally, we divide both sides by $8$ to solve for $z$ , which gives us $2z-3 = -6z + 9$ $0$ . Simplifying the fraction $2z-3 = -6z + 9$ $1$ , we get $2z-3 = -6z + 9$ $2$ or $2z-3 = -6z + 9$ $3$ . This is the solution we obtain after squaring both sides of the equation. However, we must check this solution in the original equation to ensure it is not extraneous. We substitute $2z-3 = -6z + 9$ $2$ into the original equation $2z-3 = -6z + 9$ $5$ . Substituting $2z-3 = -6z + 9$ $2$ into the left side of the original equation, we get $2z-3 = -6z + 9$ $7$ . Calculating the inside of the square root, we have $2z-3 = -6z + 9$ $8$ . Simplifying the terms inside the square root, we get $2z-3 = -6z + 9$ $9$ . Taking the square root of $6z$ $0$ , we get $6z$ $1$ .
Checking for Extraneous Solution: Next, we subtract $z^2$ from both sides to simplify the equation, which gives us $2z-3 = -6z + 9$ .Now, we add $6z$ to both sides to isolate the $z$ terms on one side, resulting in $2z + 6z - 3 = 9$ .Combining like terms, we get $8z - 3 = 9$ .Adding $3$ to both sides gives us $8z = 12$ .Finally, we divide both sides by $8$ to solve for $z$ , which gives us $2z-3 = -6z + 9$ $0$ .Simplifying the fraction $2z-3 = -6z + 9$ $1$ , we get $2z-3 = -6z + 9$ $2$ or $2z-3 = -6z + 9$ $3$ . This is the solution we obtain after squaring both sides of the equation.However, we must check this solution in the original equation to ensure it is not extraneous. We substitute $2z-3 = -6z + 9$ $2$ into the original equation $2z-3 = -6z + 9$ $5$ .Substituting $2z-3 = -6z + 9$ $2$ into the left side of the original equation, we get $2z-3 = -6z + 9$ $7$ .Calculating the inside of the square root, we have $2z-3 = -6z + 9$ $8$ .Simplifying the terms inside the square root, we get $2z-3 = -6z + 9$ $9$ .Taking the square root of $6z$ $0$ , we get $6z$ $1$ .Now we substitute $2z-3 = -6z + 9$ $2$ into the right side of the original equation, which is $6z$ $3$ .
Checking for Extraneous Solution: Next, we subtract $z^2$ from both sides to simplify the equation, which gives us $2z-3 = -6z + 9$ .Now, we add $6z$ to both sides to isolate the $z$ terms on one side, resulting in $2z + 6z - 3 = 9$ .Combining like terms, we get $8z - 3 = 9$ .Adding $3$ to both sides gives us $8z = 12$ .Finally, we divide both sides by $8$ to solve for $z$ , which gives us $2z-3 = -6z + 9$ $0$ .Simplifying the fraction $2z-3 = -6z + 9$ $1$ , we get $2z-3 = -6z + 9$ $2$ or $2z-3 = -6z + 9$ $3$ . This is the solution we obtain after squaring both sides of the equation.However, we must check this solution in the original equation to ensure it is not extraneous. We substitute $2z-3 = -6z + 9$ $2$ into the original equation $2z-3 = -6z + 9$ $5$ .Substituting $2z-3 = -6z + 9$ $2$ into the left side of the original equation, we get $2z-3 = -6z + 9$ $7$ .Calculating the inside of the square root, we have $2z-3 = -6z + 9$ $8$ .Simplifying the terms inside the square root, we get $2z-3 = -6z + 9$ $9$ .Taking the square root of $6z$ $0$ , we get $6z$ $1$ .Now we substitute $2z-3 = -6z + 9$ $2$ into the right side of the original equation, which is $6z$ $3$ .Calculating the right side, we get $6z$ $4$ .

Miku solves the equation below by first squaring both sides of the equation.\newlinez2+2z−3=z−3 \sqrt{z^{2}+2 z-3}=z-3 z2+2z−3​=z−3\newlineWhat extraneous solution does Miku obtain?\newlinez= z= z=

Miku solves the equation below by first squaring both sides of the equation. $\newline$ $\sqrt{z^{2}+2 z-3}=z-3$ $\newline$ What extraneous solution does Miku obtain? $\newline$ $z=$