Solve by completing the square. $\newline$ $y^2 - 30y + 49 = 0$ $\newline$ Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth. $\newline$ $y =$ _ or $y =$ _

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Algebra 2

Solve a quadratic equation by completing the square

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Q. Solve by completing the square.

\newline

y^2 - 30y + 49 = 0

\newline

Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.

\newline

y =

_____ or

y =

_____

Rewrite in standard form: $y^2 - 30y + 49 = 0$ $\newline$ Rewrite the equation in the form of $y^2 + by = c$ . $\newline$ Subtract $49$ from both sides. $\newline$ $y^2 - 30y + 49 - 49 = 0 - 49$ $\newline$ $y^2 - 30y = -49$
Complete the square: $y^2 - 30y = -49$ $\newline$ Choose the equation after completing the square. $\newline$ Since $\left(-\frac{30}{2}\right)^2 = 225$ , add $225$ on both sides. $\newline$ $y^2 - 30y + 225 = -49 + 225$ $\newline$ $y^2 - 30y + 225 = 176$
Factor left side: $y^2 - 30y + 225 = 176$ $\newline$ Identify the equation after factoring the left side. $\newline$ $y^2 - 30y + 225 = 176$ $\newline$ $(y - 15)^2 = 176$
Take square root: $(y - 15)^2 = 176$ $\newline$ Identify the equation after taking the square root on both sides. $\newline$ Take the square root of both sides of the equation. $\newline$ $\sqrt{(y - 15)^2} = \sqrt{176}$ $\newline$ $y - 15 = \pm\sqrt{176}$ $\newline$ $y - 15 \approx \pm13.27$
Isolate variable: We found: $\newline$ $y - 15 \approx \pm13.27$ $\newline$ Choose the equation after isolating the variable y. $\newline$ To isolate $y$ , add $15$ to both sides of the equation. $\newline$ $y - 15 + 15 \approx \pm13.27 + 15$ $\newline$ $y \approx 15 \pm 13.27$
Isolate variable: We found: $\newline$ $y - 15 \approx \pm13.27$ $\newline$ Choose the equation after isolating the variable $y$ . $\newline$ To isolate $y$ , add $15$ to both sides of the equation. $\newline$ $y - 15 + 15 \approx \pm13.27 + 15$ $\newline$ $y \approx 15 \pm 13.27$ We have: $\newline$ $y \approx 15 \pm 13.27$ $\newline$ What are the two values of $y$ ? $\newline$ $y \approx 15 + 13.27$ implies $y \approx 28.27$ . $\newline$ $y$ $0$ implies $y$ $1$ . $\newline$ Values of $y$ : $y$ $3$ , $y$ $4$