Completing the Square: Practice

The completing constant is half the coefficient of $x$, then squared.

$\left(\tfrac{6}{2}\right)^2 = 3^2 = 9$

Adding $9$ turns $x^2 + 6x$ into the perfect square $(x + 3)^2$.

Vertex form $(x - h)^2 + k$ has vertex $(h, k)$.

$y = (x - 3)^2 + 4 \ \Rightarrow\ (h, k) = (3, 4)$

The vertex's $x$-coordinate is the opposite of the sign shown inside the square.

The equation asks for every number whose square is $25$, and both $5$ and $-5$ qualify.

$x^2 = 25 \ \Rightarrow\ x = \pm\sqrt{25} = \pm 5$

The radical by itself gives only the nonnegative root, so the $\pm$ is required.

The square is already isolated, so take the square root of both sides with a $\pm$.

$x - 2 = \pm 3 \ \Rightarrow\ x = 2 \pm 3$

So $x = 5$ or $x = -1$.

A square is never negative, so the smallest value of $(x - 1)^2$ is $0$, reached at $x = 1$.

$y = (1 - 1)^2 + 2 = 2$

The minimum value is the constant $k = 2$.

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

$x^2 - 6x + 10 = (x^2 - 6x + 9) - 9 + 10 = (x - 3)^2 + 1$

The loose constants combine as $-9 + 10 = 1$.

Take the square root of both sides, keeping both signs.

$x^2 = 49 \ \Rightarrow\ x = \pm\sqrt{49} = \pm 7$

Both $7$ and $-7$ square to $49$.

The axis of symmetry passes through the vertex. Vertex form $(x - h)^2 + k$ has axis $x = h$, and here $(x + 5)^2 = (x - (-5))^2$.

$x = h = -5$

The vertex sits at $x = -5$.

Half of $-2$ is $-1$, and $(-1)^2 = 1$.

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

The constant is already exactly right, so the leftover is $0$ and the expression is itself a perfect square.

A square equals zero only when the base is zero, which happens at a single value.

$(x - 4)^2 = 0 \ \Rightarrow\ x - 4 = 0 \ \Rightarrow\ x = 4$

The two roots coincide, so there is exactly one real solution.

Take half of $14$ and square it.

$\left(\tfrac{14}{2}\right)^2 = 7^2 = 49$

Adding $49$ gives $x^2 + 14x + 49 = (x + 7)^2$.

Move the constant across, then take the root with a $\pm$.

$x^2 - 9 = 0 \ \Rightarrow\ x^2 = 9 \ \Rightarrow\ x = \pm 3$

Both $3$ and $-3$ are solutions.