Solving Quadratic Equations by Completing the Square Notes

\begin{definition term="Completing the Square"}A method for solving quadratic equations by transforming the equation into a perfect square trinomial.\end{definition}

The completing the square method is a powerful technique for solving quadratic equations. It involves transforming the quadratic expression into a perfect square trinomial, making it easier to solve.

Steps to Complete the Square

1. Ensure the coefficient of \$x^2\$ is 1. If not, divide the entire equation by the coefficient.
2. Move the constant term to the other side of the equation.
3. Add and subtract the square of half the coefficient of \$x\$ to the left side to form a perfect square trinomial.
4. Rewrite the trinomial as a squared binomial.
5. Solve for \$x\$ by taking the square root of both sides.

Example 1: Solving by Completing the Square

Solve the equation \$x^2 + 6x - 7 = 0\$ by completing the square.

2. Move the constant term to the other side:
3. $$x^2 + 6x = 7$$
5. Add and subtract the square of half the coefficient of \$x\$:
   • Half of 6 is 3, and \$3^2 = 9\$.
   • Add 9 to both sides:
   • $$x^2 + 6x + 9 = 7 + 9$$
7. Rewrite the left side as a squared binomial:
8. $$(x + 3)^2 = 16$$
10. Solve for \$x\$ by taking the square root of both sides:
11. $$x + 3 = \pm 4$$
13. Solve for \$x\$:
    • \$x + 3 = 4 \implies x = 1\$
    • \$x + 3 = -4 \implies x = -7\$

The solutions are \$x = 1\$ and \$x = -7\$.

Example 2: Completing the Square with a Leading Coefficient

Solve the equation \$2x^2 + 8x - 10 = 0\$ by completing the square.

2. Divide the entire equation by the leading coefficient (2):
3. $$x^2 + 4x - 5 = 0$$
5. Move the constant term to the other side:
6. $$x^2 + 4x = 5$$
8. Add and subtract the square of half the coefficient of \$x\$:
   • Half of 4 is 2, and \$2^2 = 4\$.
   • Add 4 to both sides:
   • $$x^2 + 4x + 4 = 5 + 4$$
10. Rewrite the left side as a squared binomial:
11. $$(x + 2)^2 = 9$$
13. Solve for \$x\$ by taking the square root of both sides:
14. $$x + 2 = \pm 3$$
16. Solve for \$x\$:
    • \$x + 2 = 3 \implies x = 1\$
    • \$x + 2 = -3 \implies x = -5\$

The solutions are \$x = 1\$ and \$x = -5\$.