Solve The Quadratic Equation By Completing The Square Worksheets [PDF]: Algebra 2 Math

How Will This Worksheet on “Solve the Quadratic Equation by Completing the Square” Benefit Your Student's Learning?

Ensure the equation is in the form $ ax^2 + bx + c = 0 $.
If the coefficient $ a $ of $ x^2 $ is not `1`, divide the entire equation by $ a $.
Move the constant term $ c $ to the other side of the equation.
Take half of the coefficient of$ x $ (which is `\frac{b}{2}`), square it `\left(\frac{b}{2}\right)^2`, and add and subtract this square inside the equation.
Rewrite the left side of the equation as a perfect square trinomial, and simplify the equation.
Take the square root of both sides of the equation, and solve for $ x $.

Solved Example

Q. Solve by completing the square.

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x^2 + 8x = 29

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

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x =

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x =

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Solution:

Write Equation Form: Write the equation in the form of $x^2 + bx = c$ . We have the equation $x^2 + 8x = 29$ .
Complete Square: Complete the square by adding $\left(\frac{b}{2}\right)^2$ to both sides of the equation. $\newline$ Since $\left(\frac{8}{2}\right)^2=16$ , we add $16$ to both sides to complete the square. $\newline$ $x^2 + 8x + 16 = 29 + 16$ $\newline$ $x^2 + 8x + 16 = 45$
Factor Left Side: Factor the left side of the equation. $\newline$ The left side is a perfect square trinomial. $\newline$ $(x + 4)^2 = 45$
Take Square Root: Take the square root of both sides of the equation. $\newline$ $\sqrt{(x + 4)^2} = \pm\sqrt{45}$ $\newline$ $x + 4 = \pm\sqrt{45}$
Solve for x: Solve for x by isolating the variable. $\newline$ Subtract $4$ from both sides of the equation. $\newline$ $x = -4 \pm \sqrt{45}$
Simplify Square Root: Simplify the square root and round to the nearest hundredth if necessary. $\newline$ $\sqrt{45}$ is approximately $6.71$ . $\newline$ $x = -4 \pm 6.71$
Find Values of x: Find the two values of x. $\newline$ $x \approx -4 + 6.71$ implies $x \approx 2.71$ . $\newline$ $x \approx -4 - 6.71$ implies $x \approx -10.71$ . $\newline$ Values of x: $2.71$ , $-10.71$