Before you read

Answer first — see what you already know.

A satellite's orbital equation $x^{2} + 10 x + 20$ must be written in vertex form. What is the result?

A
$(x + 5)^{2} - 5$
B
$(x + 5)^{2} + 5$
C
$(x + 10)^{2} - 80$
D
$(x - 5)^{2} - 5$

Check your answer

Answer:

Before you read

Answer first — see what you already know.

An antenna signal strength is $y = x^{2} - 4 x + 7$ . What is the vertex of this parabola?

A
$(2, 3)$
B
$(- 2, 15)$
C
$(4, 7)$
D
$(2, 7)$

Check your answer

Answer:

Theory

Why Complete the Square?

Vertex form of a quadratic: $y = a(x - h)^2 + k$

The vertex is $(h, k)$ — the highest or lowest point of the parabola.

Completing the square converts standard form ( $ax^2 + bx + c$ ) into vertex form. This lets you:
- Find the vertex directly
- Solve equations that don't factor nicely
- Understand the geometry of the parabola

$x^2 + 6x + 5 = 0$
$x^2 + 6x = -5$
$x^2 + 6x + 9 = 4$ add (6/2) $^{2}$ = 9
$(x+3)^2 = 4$
$x = -3 \pm 2$

Completing the square

Theory

The Method

To complete the square for $x^2 + bx$ :

1. Take half of $b$ : $\frac{b}{2}$
2. Square it: $\left(\frac{b}{2}\right)^2$
3. Add and subtract this value

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

Example 1

Rewrite $x^2 + 6x + 5$ in vertex form.

Focus on $x^2 + 6x$ . Half of $6$ is $3$ . Square: $9$ .

$x^2 + 6x + 9 - 9 + 5$

$= (x + 3)^2 - 4$

(x + 3)^2 - 4

. Vertex:

(-3, -4)

Example 2

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

$x^2 - 8x = -10$

Half of $-8$ is $-4$ . Square: $16$ . Add to both sides:

$x^2 - 8x + 16 = -10 + 16$

$(x - 4)^2 = 6$

$x - 4 = \pm\sqrt{6}$

$x = 4 \pm \sqrt{6}$

x = 4 \pm \sqrt{6}

Example 3

Rewrite $2x^2 + 12x + 7$ in vertex form.

Factor out $a$ from $x$ -terms: $2(x^2 + 6x) + 7$

Complete inside: half of $6$ is $3$ , squared is $9$ .

$2(x^2 + 6x + 9 - 9) + 7$

$= 2(x + 3)^2 - 18 + 7$

$= 2(x + 3)^2 - 11$

2(x + 3)^2 - 11

. Vertex:

(-3, -11)

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