Solve by completing the square. $\newline$ $v^2 - 28v + 17 = 0$ $\newline$ Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth. $\newline$ $v =$ _ or $v =$ _

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Algebra 2

Solve a quadratic equation by completing the square

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Q. Solve by completing the square.

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v^2 - 28v + 17 = 0

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

\newline

v =

_____ or

v =

_____

Set up for completing the square: Write the equation in the form of $v^2 + bv = c$ . The given equation is already in this form: $v^2 - 28v + 17 = 0$ . Subtract $17$ from both sides to set the equation up for completing the square. $v^2 - 28v = -17$
Add number to complete square: Find the number to add to both sides to complete the square. $\newline$ To complete the square, we need to add $(b/2)^2$ to both sides, where $b$ is the coefficient of $v$ . $\newline$ In this case, $b = -28$ , so $(b/2)^2 = (-28/2)^2 = (-14)^2 = 196$ . $\newline$ Add $196$ to both sides of the equation. $\newline$ $v^2 − 28v + 196 = -17 + 196$ $\newline$ $v^2 − 28v + 196 = 179$
Factor left side: Factor the left side of the equation. $\newline$ The left side of the equation is now a perfect square trinomial. $\newline$ $(v - 14)^2 = 179$
Take square root: Take the square root of both sides. $\newline$ To solve for $v$ , take the square root of both sides of the equation. $\newline$ $\sqrt{(v - 14)^2} = \pm\sqrt{179}$ $\newline$ $v - 14 = \pm\sqrt{179}$
Solve for $v$ : Solve for $v$ . $\newline$ Add $14$ to both sides of the equation to isolate $v$ . $\newline$ $v = 14 \pm \sqrt{179}$ $\newline$ Since $\sqrt{179}$ is not a perfect square, we can approximate it to the nearest hundredth. $\newline$ $\sqrt{179} \approx 13.38$ $\newline$ $v \approx 14 \pm 13.38$
Find two values of $v$ : Find the two values of $v$ . $v \approx 14 + 13.38$ implies $v \approx 27.38$ . $v \approx 14 - 13.38$ implies $v \approx 0.62$ .Values of $v$ : $27.38$ , $0.62$