Completing the Square: Practice

Half of $5$ is $\tfrac{5}{2}$, and squaring gives the completing constant.

$\left(\tfrac{5}{2}\right)^2 = \tfrac{25}{4}$

Then $x^2 + 5x + \tfrac{25}{4} = \left(x + \tfrac{5}{2}\right)^2$.

Complete the square: half of $-4$ is $-2$, and $(-2)^2 = 4$.

$x^2 - 4x - 1 = (x - 2)^2 - 4 - 1 = (x - 2)^2 - 5$

Set to zero and isolate: $(x - 2)^2 = 5$, so $x - 2 = \pm\sqrt{5}$ and $x = 2 \pm \sqrt{5}$.

Complete the square: half of $4$ is $2$, and $2^2 = 4$.

$x^2 + 4x + 7 = (x + 2)^2 - 4 + 7 = (x + 2)^2 + 3$

Setting this to zero gives $(x + 2)^2 = -3$. A real square cannot be negative, so there is no real solution.

Half of $6$ is $3$, and $3^2 = 9$.

$x^2 + 6x + 4 = (x + 3)^2 - 9 + 4 = (x + 3)^2 - 5$

Set to zero: $(x + 3)^2 = 5$, so $x + 3 = \pm\sqrt{5}$ and $x = -3 \pm \sqrt{5}$.

Factor $3$ from the first two terms, then complete the square.

$3x^2 - 12x + 5 = 3(x^2 - 4x) + 5 = 3(x - 2)^2 - 12 + 5 = 3(x - 2)^2 - 7$

Vertex form gives vertex $(2, -7)$.

Complete the square: half of $-4$ is $-2$, and $(-2)^2 = 4$.

$x^2 - 4x + 9 = (x - 2)^2 - 4 + 9 = (x - 2)^2 + 5$

The leading coefficient is positive, so the minimum value is $5$ and the range is $y \ge 5$.

Half of $-7$ is $-\tfrac{7}{2}$, and squaring removes the sign.

$\left(-\tfrac{7}{2}\right)^2 = \tfrac{49}{4}$

Then $x^2 - 7x + \tfrac{49}{4} = \left(x - \tfrac{7}{2}\right)^2$.

The trinomial is a perfect square exactly when the constant equals the square of half the middle coefficient.

$c = \left(\tfrac{10}{2}\right)^2 = 25$

Then $x^2 + 10x + 25 = (x + 5)^2$.

The left side is already a perfect square, since half of $-6$ squared is $9$.

$x^2 - 6x + 9 = (x - 3)^2 = 0 \ \Rightarrow\ x = 3$

The two roots coincide, so there is exactly one solution, $x = 3$.

Complete the square to reach vertex form.

$2x^2 + 8x + 1 = 2(x^2 + 4x) + 1 = 2(x + 2)^2 - 8 + 1 = 2(x + 2)^2 - 7$

The axis of symmetry passes through the vertex $x$-coordinate, $x = -2$.

Half of $2$ is $1$, and $1^2 = 1$.

$x^2 + 2x - 5 = (x + 1)^2 - 1 - 5 = (x + 1)^2 - 6$

Set to zero: $(x + 1)^2 = 6$, so $x + 1 = \pm\sqrt{6}$ and $x = -1 \pm \sqrt{6}$.

Complete the square: half of $8$ is $4$, and $4^2 = 16$.

$x^2 + 8x + 20 = (x + 4)^2 - 16 + 20 = (x + 4)^2 + 4$

The minimum value is $4 > 0$, so $y$ is always positive and the graph never crosses the $x$-axis.