How to Complete the Square (Step by Step)

A perfect square trinomial has this pattern:

$$x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2$$

So to complete the square on $x^2 + bx$, take half of $b$ and square it, then add that number.

Example: Complete the square on $x^2 + 6x$.

1. Half of 6 is 3.
2. Square it: $3^2 = 9$.
3. Add it: $x^2 + 6x + 9 = (x + 3)^2$.

Example: Complete the square on $x^2 - 10x$.

1. Half of $-10$ is $-5$, and $(-5)^2 = 25$.
2. $x^2 - 10x + 25 = (x - 5)^2$.

The number inside the parentheses is always half of $b$, with the same sign as $b$.

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

1. Move the constant to the right: $x^2 + 6x = 7$.
2. Half of 6 is 3, and $3^2 = 9$. Add 9 to both sides: $x^2 + 6x + 9 = 16$.
3. Write the left side as a square: $(x + 3)^2 = 16$.
4. Take the square root of both sides, with $\pm$: $x + 3 = \pm 4$.
5. Solve each case: $x = -3 + 4 = 1$ or $x = -3 - 4 = -7$.

Check: $1 + 6 - 7 = 0$ and $49 - 42 - 7 = 0$. Both work.

An odd $b$ just means you work with fractions. Do not round.

Example: Solve $x^2 + 5x - 2 = 0$.

1. Move the constant: $x^2 + 5x = 2$.
2. Half of 5 is $\frac{5}{2}$. Square it: $\frac{25}{4}$.
3. Add to both sides: $x^2 + 5x + \frac{25}{4} = 2 + \frac{25}{4} = \frac{8}{4} + \frac{25}{4} = \frac{33}{4}$.
4. Factor the left side: $\left(x + \frac{5}{2}\right)^2 = \frac{33}{4}$.
5. Square root both sides: $x + \frac{5}{2} = \pm\frac{\sqrt{33}}{2}$.
6. Solve: $x = \frac{-5 \pm \sqrt{33}}{2}$.

The rule only works when the coefficient of $x^2$ is 1. If it is not, divide every term by $a$ first.

Example: Solve $2x^2 + 8x - 3 = 0$.

1. Divide every term by 2: $x^2 + 4x - \frac{3}{2} = 0$.
2. Move the constant: $x^2 + 4x = \frac{3}{2}$.
3. Half of 4 is 2, and $2^2 = 4$. Add 4 to both sides: $x^2 + 4x + 4 = \frac{3}{2} + 4 = \frac{11}{2}$.
4. Factor: $(x + 2)^2 = \frac{11}{2}$.
5. Square root: $x + 2 = \pm\sqrt{\frac{11}{2}} = \pm\frac{\sqrt{22}}{2}$.
6. Solve: $x = -2 \pm \frac{\sqrt{22}}{2}$.

To simplify $\sqrt{\frac{11}{2}}$, multiply the top and bottom by 2 inside the root: $\sqrt{\frac{22}{4}} = \frac{\sqrt{22}}{2}$.

Vertex form is $y = a(x - h)^2 + k$, where the vertex is $(h, k)$. Completing the square gets you there. Since this is an expression (not an equation), you add and subtract the same number instead of adding to both sides.

Example 1: Write $y = x^2 - 4x + 7$ in vertex form.

1. Half of $-4$ is $-2$, and $(-2)^2 = 4$.
2. Add and subtract 4: $y = (x^2 - 4x + 4) - 4 + 7$.
3. Factor: $y = (x - 2)^2 + 3$.
4. The vertex is $(2, 3)$.

Example 2: Write $y = 3x^2 + 12x + 5$ in vertex form.

1. Factor 3 out of the $x$ terms only: $y = 3(x^2 + 4x) + 5$.
2. Inside, half of 4 is 2, and $2^2 = 4$. Add and subtract 4 inside: $y = 3(x^2 + 4x + 4 - 4) + 5$.
3. Move the $-4$ outside. It gets multiplied by 3 on the way out: $y = 3(x^2 + 4x + 4) - 12 + 5$.
4. Simplify: $y = 3(x + 2)^2 - 7$.
5. The vertex is $(-2, -7)$. Check: $3(4) + 12(-2) + 5 = 12 - 24 + 5 = -7$.

If completing the square leaves a negative number on the right, there are no real solutions.

Example: Solve $x^2 + 2x + 5 = 0$.

1. $x^2 + 2x = -5$.
2. Add 1: $(x + 1)^2 = -4$.
3. No real number squared is negative, so there are no real solutions. With complex numbers, $x = -1 \pm 2i$.

Where the quadratic formula comes from. Complete the square on $ax^2 + bx + c = 0$:

1. Divide by $a$ and move the constant: $x^2 + \frac{b}{a}x = -\frac{c}{a}$.
2. Add $\left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}$ to both sides: $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$.
3. Square root and solve:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Paste a quadratic into the VeilGPT solver to check your steps.