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$Rewrite the equation by completing the square. {:[x^(2)-x-20=0],[(x+◻)^(2)=◻]:}$

Rewrite the equation by completing the square. $\newline$ $x^2 - x - 20 = 0$ $\newline$ $(x+\square)^2 = \square$

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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 - x - 20 = 0

\newline

(x+\square)^2 = \square

Rearrange the equation: $x^2 - x - 20 = 0$ $\newline$ Rearrange the equation to the form $x^2 + bx$ by moving the constant term to the right side of the equation. $\newline$ $x^2 - x = 20$
Complete the square: $x^2 - x = 20$ $\newline$ To complete the square, add the square of half the coefficient of $x$ to both sides of the equation. The coefficient of $x$ is $-1$ , so half of it is $-\frac{1}{2}$ , and the square of $-\frac{1}{2}$ is $\frac{1}{4}$ . $\newline$ $x^2 - x + \left(\frac{1}{4}\right) = 20 + \left(\frac{1}{4}\right)$ $\newline$ $x^2 - x + \left(\frac{1}{4}\right) = \frac{80}{4} + \frac{1}{4}$ $\newline$ $x^2 - x + \left(\frac{1}{4}\right) = \frac{81}{4}$
Simplify the equation: $x^2 - x + \left(\frac{1}{4}\right) = \frac{81}{4}$ $\newline$ The left side of the equation is now a perfect square, and can be written as $(x - \frac{1}{2})^2$ . $\newline$ $(x - \frac{1}{2})^2 = \frac{81}{4}$
Write in completed square form: $(x - \frac{1}{2})^2 = \frac{81}{4}$ $\newline$ This is the completed square form of the equation.

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