Completing the Square: Additional Practice (Set 1)

Completing the square gives $(x - 2)^2 + (c - 4) = 0$, that is $(x - 2)^2 = 4 - c$. Two distinct real roots need a positive right side.

$4 - c > 0 \ \Rightarrow\ c < 4$

When $c = 4$ the roots coincide, and when $c > 4$ there is no real solution.

Half of $7$ is $\tfrac{7}{2}$, and $\left(\tfrac{7}{2}\right)^2 = \tfrac{49}{4}$.

$x^2 + 7x + 10 = \left(x + \tfrac{7}{2}\right)^2 - \tfrac{49}{4} + 10 = \left(x + \tfrac{7}{2}\right)^2 - \tfrac{9}{4}$

The constants combine as $-\tfrac{49}{4} + \tfrac{40}{4} = -\tfrac{9}{4}$.

Factor $2$ and complete the square: $2x^2 - 8x + 5 = 2(x - 2)^2 - 3$.

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

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

Complete the square: half of $2$ is $1$, and $1^2 = 1$.

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

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

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

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

The leading coefficient is negative, so the maximum value is $7$, at $x = 2$.

Half of $5$ is $\tfrac{5}{2}$, and $\left(\tfrac{5}{2}\right)^2 = \tfrac{25}{4}$.

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

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

Complete the square: $x^2 - 4x + c = (x - 2)^2 + (c - 4)$. The minimum value is the constant $c - 4$.

$c - 4 = 1 \ \Rightarrow\ c = 5$

Then $f(x) = (x - 2)^2 + 1$, with minimum $1$ at $x = 2$.

Divide by $3$ to isolate the square, then take the root with a $\pm$.

$(x + 2)^2 = 4 \ \Rightarrow\ x + 2 = \pm 2$

So $x = 0$ or $x = -4$.

Half of $1$ is $\tfrac{1}{2}$, and $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$.

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

Setting this to zero needs $\left(x + \tfrac{1}{2}\right)^2 = -\tfrac{3}{4}$, impossible for a real square, so there is no real solution.

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

$x^2 - 10x + 18 = (x - 5)^2 - 25 + 18 = (x - 5)^2 - 7$

Set to zero: $(x - 5)^2 = 7$, so $x - 5 = \pm\sqrt{7}$ and $x = 5 \pm \sqrt{7}$.

Complete the square and solve.

$x^2 + 6x + 5 = (x + 3)^2 - 4 = 0 \ \Rightarrow\ x + 3 = \pm 2$

The roots are $x = -1$ and $x = -5$, whose sum is $-6$.

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

$-2x^2 - 4x + 1 = -2(x^2 + 2x) + 1 = -2\big((x + 1)^2 - 1\big) + 1$

Distribute the $-2$: $-2(x + 1)^2 + 2 + 1 = -2(x + 1)^2 + 3$.