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$Rewrite the equation by completing the square. {:[x^(2)-4x-32=0],[(x+◻)^(2)=◻]:}$

Rewrite the equation by completing the square. $\newline$ $x^2-4x-32=0$ $\newline$ $(x+\square)^2=\square$

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Solve a quadratic equation by completing the square

Full solution

Q. Rewrite the equation by completing the square.

\newline

x^2-4x-32=0

\newline

(x+\square)^2=\square

Given equation: Start with the given equation $x^2 - 4x - 32 = 0$ . $\newline$ To complete the square, we need to move the constant term to the other side of the equation. $\newline$ Add $32$ to both sides. $\newline$ $x^2 - 4x - 32 + 32 = 0 + 32$ $\newline$ $x^2 - 4x = 32$
Move constant term: To complete the square, we need to find a number that, when added to $x^2 - 4x$ , will make it a perfect square trinomial. $\newline$ The number we need to add is $(\frac{b}{2})^2$ , where $b$ is the coefficient of $x$ . $\newline$ In this case, $b = -4$ , so $(\frac{b}{2})^2 = (\frac{-4}{2})^2 = (-2)^2 = 4$ . $\newline$ Add $4$ to both sides of the equation. $\newline$ $x^2 - 4x + 4 = 32 + 4$ $\newline$ $x^2 - 4x + 4 = 36$
Add number to complete the square: Now the left side of the equation is a perfect square trinomial. $\newline$ Factor the left side. $\newline$ $(x - 2)^2 = 36$
Factor the left side: Write the completed square equation. $\newline$ \[{[x^\(2\) - \(4\)x - \(32\) = \(0\)],[(x - \(2\))^\(2\) = \(36\)]:}

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