Completing the Square: Practice

Completing the square gives $(x - 4)^2 + (k - 16) = 0$, that is $(x - 4)^2 = 16 - k$. There is exactly one solution when the right side is $0$.

$16 - k = 0 \ \Rightarrow\ k = 16$

Then $(x - 4)^2 = 0$, a single repeated root at $x = 4$.

The vertex lies on the $y$-axis when its $x$-coordinate is $0$. Completing the square puts the vertex at $x = -\tfrac{b}{2}$.

$-\tfrac{b}{2} = 0 \ \Rightarrow\ b = 0$

With $b = 0$ the function is $x^2 + 10$, whose vertex $(0, 10)$ sits on the axis.

Completing the square gives $(x + 3)^2 + (c - 9) = 0$, that is $(x + 3)^2 = 9 - c$. A real square is never negative, so there is no real solution when the right side is negative.

$9 - c < 0 \ \Rightarrow\ c > 9$

For $c > 9$ the parabola stays entirely above the $x$-axis.

Factor $-1$ from the first two terms, then complete the square.

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

This simplifies to $-(x - 3)^2 + 5$.

Factor $3$ and complete the square: $3x^2 - 12x + 4 = 3(x - 2)^2 - 8$.

$3(x - 2)^2 - 8 = 0 \ \Rightarrow\ (x - 2)^2 = \tfrac{8}{3}$

Then $x - 2 = \pm\sqrt{\tfrac{8}{3}} = \pm\dfrac{2\sqrt{2}}{\sqrt{3}} = \pm\dfrac{2\sqrt{6}}{3}$, so $x = 2 \pm \dfrac{2\sqrt{6}}{3}$.

Factor $4$ from the first two terms and complete the square.

$4x^2 + 4x + 1 = 4\left(x^2 + x\right) + 1 = 4\left(x + \tfrac{1}{2}\right)^2 - 1 + 1 = 4\left(x + \tfrac{1}{2}\right)^2$

Setting this to zero forces $x = -\tfrac{1}{2}$, a single repeated root, so there is exactly one real solution.

Half of $8$ is $4$, and $4^2 = 16$.

$x^2 + 8x + 11 = (x + 4)^2 - 16 + 11 = (x + 4)^2 - 5$

Set to zero: $(x + 4)^2 = 5$, so $x + 4 = \pm\sqrt{5}$ and $x = -4 \pm \sqrt{5}$.

A perfect square needs the constant to equal the square of half the middle coefficient: $\left(\tfrac{b}{2}\right)^2 = 9$.

$\tfrac{b}{2} = \pm 3 \ \Rightarrow\ b = \pm 6$

Both work: $x^2 + 6x + 9 = (x + 3)^2$ and $x^2 - 6x + 9 = (x - 3)^2$.

Isolate the square first.

$2(x - 3)^2 - 10 = 0 \ \Rightarrow\ (x - 3)^2 = 5$

Then $x - 3 = \pm\sqrt{5}$, so $x = 3 \pm \sqrt{5}$.

The expression is already a perfect square.

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

The graph touches the $x$-axis at the single point $x = -2$, where the vertex sits.

Half of $3$ is $\tfrac{3}{2}$, and $\left(\tfrac{3}{2}\right)^2 = \tfrac{9}{4}$.

$x^2 + 3x + 1 = \left(x + \tfrac{3}{2}\right)^2 - \tfrac{9}{4} + 1 = \left(x + \tfrac{3}{2}\right)^2 - \tfrac{5}{4}$

So $\left(x + \tfrac{3}{2}\right)^2 = \tfrac{5}{4}$, giving $x = -\tfrac{3}{2} \pm \dfrac{\sqrt{5}}{2} = \dfrac{-3 \pm \sqrt{5}}{2}$.

Factor $2$ from the first two terms, then complete the square.

$2x^2 - 8x + 3 = 2(x^2 - 4x) + 3 = 2(x - 2)^2 - 8 + 3 = 2(x - 2)^2 - 5$

The leading coefficient is positive, so the minimum value is $-5$, at $x = 2$.