Lea solves the equation below by first squaring both sides of the equation. 3+2y=sqrt(-y) What extraneous solution does Lea obtain? y=

Perform Squaring: Now, let's perform the squaring on both sides. (3 + 2y)^2 = -y 9 + 12y + 4y^2 = -yMove Terms: Next, we'll move all terms to one side to set the equation to zero. 4y^2 + 12y + y + 9 = 0 4y^2 + 13y + 9 = 0Solve Quadratic Equation: Now, we need to solve the quadratic equation for y. We can use the quadratic formula, y = (-b ± sqrt(b^2 - 4ac)) / (2a), where a = 4, b = 13, and c = 9.Calculate Discriminant: Let's calculate the discriminant (b^2 - 4ac) first. Discriminant = 13^2 - 4(4)(9) Discriminant = 169 - 144 Discriminant = 25Find Solutions: Since the discriminant is positive, we have two real solutions. Let's find them using the quadratic formula. y = (-13 ± sqrt(25)) / (2 * 4) y = (-13 ± 5) / 8Check Validity: Now, let's find the two solutions. First solution: y = (-13 + 5) / 8 y = -8 / 8 y = -1 Second solution: y = (-13 - 5) / 8 y = -18 / 8 y = -2.25We have two potential solutions, y = -1 and y = -2.25. However, we need to check these solutions in the original equation to see if any of them is extraneous. Original equation: 3 + 2y = sqrt(-y)Let's substitute y = -1 into the original equation. 3 + 2(-1) = sqrt(-(-1)) 3 - 2 = sqrt(1) 1 = 1 This solution is valid.Now, let's substitute y = -2.25 into the original equation. 3 + 2(-2.25) = sqrt(-(-2.25)) 3 - 4.5 = sqrt(2.25) -1.5 = 1.5 This is not true, so y = -2.25 is an extraneous solution.

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Lea solves the equation below by first squaring both sides of the equation.

3+2y=sqrt(-y)
What extraneous solution does Lea obtain?

y=

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

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

\newline

3+2 y=\sqrt{-y}

\newline

What extraneous solution does Lea obtain?

\newline

y=

Perform Squaring: Now, let's perform the squaring on both sides. $\newline$ $(3 + 2y)^2 = -y$ $\newline$ $9 + 12y + 4y^2 = -y$
Move Terms: Next, we'll move all terms to one side to set the equation to zero. $\newline$ $4y^2 + 12y + y + 9 = 0$ $\newline$ $4y^2 + 13y + 9 = 0$
Solve Quadratic Equation: Now, we need to solve the quadratic equation for $y$ . We can use the quadratic formula, $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 4$ , $b = 13$ , and $c = 9$ .
Calculate Discriminant: Let's calculate the discriminant $b^2 - 4ac$ first. $\newline$ Discriminant = $13^2 - 4(4)(9)$ $\newline$ Discriminant = $169 - 144$ $\newline$ Discriminant = $25$
Find Solutions: Since the discriminant is positive, we have two real solutions. Let's find them using the quadratic formula. $\newline$ $y = \frac{-13 \pm \sqrt{25}}{2 \times 4}$ $\newline$ $y = \frac{-13 \pm 5}{8}$
Check Validity: Now, let's find the two solutions. $\newline$ First solution: $y = (-13 + 5) / 8$ $\newline$ $y = -8 / 8$ $\newline$ $y = -1$ $\newline$ Second solution: $y = (-13 - 5) / 8$ $\newline$ $y = -18 / 8$ $\newline$ $y = -2.25$
Check Validity: Now, let's find the two solutions. $\newline$ First solution: $y = (-13 + 5) / 8$ $\newline$ $y = -8 / 8$ $\newline$ $y = -1$ $\newline$ Second solution: $y = (-13 - 5) / 8$ $\newline$ $y = -18 / 8$ $\newline$ $y = -2.25$ We have two potential solutions, $y = -1$ and $y = -2.25$ . However, we need to check these solutions in the original equation to see if any of them is extraneous. $\newline$ Original equation: $3 + 2y = \sqrt{-y}$
Check Validity: Now, let's find the two solutions. $\newline$ First solution: $y = (-13 + 5) / 8$ $\newline$ $y = -8 / 8$ $\newline$ $y = -1$ $\newline$ Second solution: $y = (-13 - 5) / 8$ $\newline$ $y = -18 / 8$ $\newline$ $y = -2.25$ We have two potential solutions, $y = -1$ and $y = -2.25$ . However, we need to check these solutions in the original equation to see if any of them is extraneous. $\newline$ Original equation: $3 + 2y = \sqrt{-y}$ Let's substitute $y = -1$ into the original equation. $\newline$ $y = -8 / 8$ $0$ $\newline$ $y = -8 / 8$ $1$ $\newline$ $y = -8 / 8$ $2$ $\newline$ This solution is valid.
Check Validity: Now, let's find the two solutions. $\newline$ First solution: $y = (-13 + 5) / 8$ $\newline$ $y = -8 / 8$ $\newline$ $y = -1$ $\newline$ Second solution: $y = (-13 - 5) / 8$ $\newline$ $y = -18 / 8$ $\newline$ $y = -2.25$ We have two potential solutions, $y = -1$ and $y = -2.25$ . However, we need to check these solutions in the original equation to see if any of them is extraneous. $\newline$ Original equation: $3 + 2y = \sqrt{-y}$ Let's substitute $y = -1$ into the original equation. $\newline$ $y = -8 / 8$ $0$ $\newline$ $y = -8 / 8$ $1$ $\newline$ $y = -8 / 8$ $2$ $\newline$ This solution is valid.Now, let's substitute $y = -2.25$ into the original equation. $\newline$ $y = -8 / 8$ $4$ $\newline$ $y = -8 / 8$ $5$ $\newline$ $y = -8 / 8$ $6$ $\newline$ This is not true, so $y = -2.25$ is an extraneous solution.