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

- Clarifies completing the square method for quadratic equations.
- Develops critical thinking and algebraic skills.
- Enhances understanding of quadratic graphs.
- Strengthens equation manipulation abilities.
- Applies math concepts to real-world scenarios.
- Prepares for advanced math topics.
- Encourages self-paced learning.

**How to Solve the Quadratic Equation by Completing the Square?**

- 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.$\newline$$x^2 + 8x = 29$$\newline$Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.$\newline$$x =$_____ or $x =$_____

**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: $2.71$, $-10.71$