Completing the Square GCSE Questions

Answer $(x + 3)^2 - 9$

1. $$x^2 + 6x = (x + 3)^2 - 3^2$$Halve 6 to get 3
2. $$= (x + 3)^2 - 9$$

• B1 $(x + 3)^2$
• B1 $-9$

Answer $(x + 5)^2 - 18$

1. $$x^2 + 10x + 7 = (x + 5)^2 - 25 + 7$$Halve 10 to get 5; subtract $5^2$
2. $$= (x + 5)^2 - 18$$

• B1 $(x + 5)^2$
• B1 $-18$

Answer $(x - 4)^2 - 13$

1. $$x^2 - 8x + 3 = (x - 4)^2 - 16 + 3$$Half of $-8$ is $-4$
2. $$= (x - 4)^2 - 13$$

• B1 $(x - 4)^2$
• B1 $-13$

Answer $(x - 2)^2 - 9$

1. $$x^2 - 4x - 5 = (x - 2)^2 - 4 - 5$$Half of $-4$ is $-2$; $(-2)^2 = 4$
2. $$= (x - 2)^2 - 9$$

• B1 $(x - 2)^2$
• B1 $-9$

Answer $a = 6,\ b = 4$

1. $$x^2 + 12x + 40 = (x + 6)^2 - 36 + 40$$Halve 12 to get 6
2. $$= (x + 6)^2 + 4$$
3. $$a = 6,\quad b = 4$$

• B1 $a = 6$
• B1 $b = 4$

Answer $\left(x + \frac{5}{2}\right)^2 - \frac{17}{4}$

1. $$x^2 + 5x + 2 = \left(x + \tfrac{5}{2}\right)^2 - \tfrac{25}{4} + 2$$Half of 5 is $\tfrac{5}{2}$
2. $$= \left(x + \tfrac{5}{2}\right)^2 - \tfrac{25}{4} + \tfrac{8}{4}$$Write 2 as quarters
3. $$= \left(x + \tfrac{5}{2}\right)^2 - \tfrac{17}{4}$$

• B1 $\left(x + \frac{5}{2}\right)^2$ — allow $(x + 2.5)^2$
• B1 $-\frac{17}{4}$ — allow $-4.25$

Answer $\left(x - \frac{3}{2}\right)^2 - \frac{13}{4}$

1. $$x^2 - 3x - 1 = \left(x - \tfrac{3}{2}\right)^2 - \tfrac{9}{4} - 1$$Half of $-3$ is $-\tfrac{3}{2}$
2. $$= \left(x - \tfrac{3}{2}\right)^2 - \tfrac{13}{4}$$

• B1 $\left(x - \frac{3}{2}\right)^2$ — allow $(x - 1.5)^2$
• B1 $-\frac{13}{4}$ — allow $-3.25$

Answer $x = 1$ or $x = -7$

1. $$(x + 3)^2 - 9 - 7 = 0$$Complete the square
2. $$(x + 3)^2 = 16$$
3. $$x + 3 = \pm 4$$Square root — keep both signs
4. $$x = 1 \quad\text{or}\quad x = -7$$

• M1 $(x + 3)^2 - 16 = 0$
• M1 $x + 3 = \pm 4$
• A1 $x = 1$ and $x = -7$

Answer $x = 2 \pm \sqrt{5}$

1. $$(x - 2)^2 - 4 - 1 = 0$$
2. $$(x - 2)^2 = 5$$
3. $$x - 2 = \pm\sqrt{5}$$
4. $$x = 2 \pm \sqrt{5}$$

• M1 $(x - 2)^2 - 5 = 0$
• M1 $x - 2 = \pm\sqrt{5}$
• A1 $x = 2 \pm \sqrt{5}$

Answer $(3,\ 2)$

1. $$x^2 - 6x + 11 = (x - 3)^2 - 9 + 11$$
2. $$= (x - 3)^2 + 2$$
3. $$\text{Turning point} = (3,\ 2)$$$(x - a)^2 + b$ has turning point $(a, b)$

• M1 $(x - 3)^2 \ldots$
• A1 $(x - 3)^2 + 2$
• A1 $(3,\ 2)$ — ft their completed square

Answer $2(x + 3)^2 - 13$

1. $$2x^2 + 12x + 5 = 2(x^2 + 6x) + 5$$Factor 2 out of the $x$ terms
2. $$= 2\left[(x + 3)^2 - 9\right] + 5$$
3. $$= 2(x + 3)^2 - 18 + 5$$Multiply the $-9$ by 2
4. $$= 2(x + 3)^2 - 13$$

• M1 $2(x^2 + 6x) + 5$
• M1 $2[(x + 3)^2 - 9] + 5$
• A1 $2(x + 3)^2 - 13$

Answer $x = -0.39$ or $x = -7.61$

1. $$(x + 4)^2 - 16 + 3 = 0$$
2. $$(x + 4)^2 = 13$$
3. $$x = -4 \pm \sqrt{13}$$
4. $$x = -0.39 \quad\text{or}\quad x = -7.61$$$\sqrt{13} = 3.6055\ldots$

• M1 $(x + 4)^2 - 13 = 0$
• M1 $x = -4 \pm \sqrt{13}$
• A1 $-0.39$ and $-7.61$

Answer $(x - 5)^2 + 3$

1. $$x^2 - 10x + 28 = (x - 5)^2 - 25 + 28$$
2. $$= (x - 5)^2 + 3$$

• B1 $(x - 5)^2$
• B1 $+3$

Answer $(x - 5)^2 \ge 0$, so $(x - 5)^2 + 3 \ge 3 > 0$.

1. $$(x - 5)^2 \ge 0$$A square is never negative
2. $$(x - 5)^2 + 3 \ge 3 > 0$$

• B1 States a square is never negative, so the expression is at least 3

Answer $(x + 2)^2 - 16$

1. $$x^2 + 4x - 12 = (x + 2)^2 - 4 - 12$$
2. $$= (x + 2)^2 - 16$$

• B1 $(x + 2)^2$
• B1 $-16$

Answer $(-2,\ -16)$

1. $$(x + 2)^2 - 16 \;\Rightarrow\; (-2,\ -16)$$Change the sign inside the bracket

• B1 $(-2,\ -16)$ — ft their (a)

Answer $x = 2$ and $x = -6$

1. $$(x + 2)^2 = 16$$
2. $$x + 2 = \pm 4$$
3. $$x = 2 \quad\text{or}\quad x = -6$$

• M1 $x + 2 = \pm 4$
• A1 $x = 2$ and $x = -6$

Answer $3(x - 2)^2 - 5$

1. $$3x^2 - 12x + 7 = 3(x^2 - 4x) + 7$$
2. $$= 3\left[(x - 2)^2 - 4\right] + 7$$
3. $$= 3(x - 2)^2 - 12 + 7$$
4. $$= 3(x - 2)^2 - 5$$

• M1 $3(x^2 - 4x) + 7$
• M1 $3[(x - 2)^2 - 4] + 7$
• A1 $3(x - 2)^2 - 5$

Answer $x = 2 \pm \frac{\sqrt{15}}{3}$

1. $$3(x - 2)^2 = 5$$
2. $$(x - 2)^2 = \tfrac{5}{3}$$
3. $$x = 2 \pm \sqrt{\tfrac{5}{3}} = 2 \pm \tfrac{\sqrt{15}}{3}$$Rationalise: $\sqrt{\tfrac{5}{3}} = \tfrac{\sqrt{5}\sqrt{3}}{3}$

• M1 $(x - 2)^2 = \frac{5}{3}$
• A1 $2 \pm \frac{\sqrt{15}}{3}$ — allow $2 \pm \sqrt{\frac{5}{3}}$ for M1 only

Answer $14 - (x + 3)^2$

1. $$5 - 6x - x^2 = 5 - (x^2 + 6x)$$Take out $-1$ from the $x$ terms
2. $$= 5 - \left[(x + 3)^2 - 9\right]$$
3. $$= 14 - (x + 3)^2$$$-(-9) = +9$

• M1 $5 - (x^2 + 6x)$
• M1 $5 - [(x + 3)^2 - 9]$
• A1 $14 - (x + 3)^2$

Answer Maximum value $14$ when $x = -3$

1. $$(x + 3)^2 \ge 0 \;\Rightarrow\; 14 - (x + 3)^2 \le 14$$
2. $$\text{Max} = 14 \text{ when } x + 3 = 0,\ x = -3$$

• B1 Maximum $14$
• B1 $x = -3$

Answer $p = 10,\ q = 5$

1. $$(x + q)^2 - 5 = x^2 + 2qx + q^2 - 5$$Expand the right-hand side
2. $$q^2 - 5 = 20 \;\Rightarrow\; q^2 = 25 \;\Rightarrow\; q = 5$$Compare constants; $q > 0$
3. $$p = 2q = 10$$Compare $x$ coefficients

• M1 Expands $(x + q)^2 - 5 = x^2 + 2qx + q^2 - 5$
• M1 $q^2 - 5 = 20$
• A1 $q = 5$ and $p = 10$

Answer $4x^2 - 12x + 10 = 4\left(x - \frac{3}{2}\right)^2 + 1 \ge 1 > 0$

1. $$4x^2 - 12x + 10 = 4(x^2 - 3x) + 10$$
2. $$= 4\left[\left(x - \tfrac{3}{2}\right)^2 - \tfrac{9}{4}\right] + 10$$
3. $$= 4\left(x - \tfrac{3}{2}\right)^2 + 1$$
4. $$4\left(x - \tfrac{3}{2}\right)^2 \ge 0 \;\Rightarrow\; \text{expression} \ge 1 > 0$$State the reason — it earns the final mark

• M1 $4\left(x^2 - 3x\right) + 10$ or $(2x - 3)^2 \ldots$
• A1 $4\left(x - \frac{3}{2}\right)^2 + 1$ or $(2x - 3)^2 + 1$
• A1 Conclusion: square $\ge 0$, so expression $\ge 1 > 0$

Answer Minimum at $(2,\ 3)$; $f(x) \ge 3$

1. $$2x^2 - 8x + 11 = 2(x^2 - 4x) + 11$$
2. $$= 2\left[(x - 2)^2 - 4\right] + 11 = 2(x - 2)^2 + 3$$
3. $$\text{Turning point } (2,\ 3)$$Positive $x^2$ coefficient, so it is a minimum
4. $$f(x) \ge 3$$

• M1 $2(x^2 - 4x) + 11$
• A1 $2(x - 2)^2 + 3$
• B1 $(2,\ 3)$, minimum — ft their completed square
• B1 $f(x) \ge 3$

Answer $x = \frac{-3 \pm \sqrt{15}}{2}$

1. $$2\left(x^2 + 3x\right) - 3 = 0$$
2. $$2\left[\left(x + \tfrac{3}{2}\right)^2 - \tfrac{9}{4}\right] - 3 = 0$$
3. $$2\left(x + \tfrac{3}{2}\right)^2 = \tfrac{15}{2} \;\Rightarrow\; \left(x + \tfrac{3}{2}\right)^2 = \tfrac{15}{4}$$
4. $$x + \tfrac{3}{2} = \pm\tfrac{\sqrt{15}}{2}$$
5. $$x = \dfrac{-3 \pm \sqrt{15}}{2}$$

• M1 $2\left(x + \frac{3}{2}\right)^2 - \frac{9}{2} - 3 = 0$
• M1 $\left(x + \frac{3}{2}\right)^2 = \frac{15}{4}$
• M1 $x + \frac{3}{2} = \pm\frac{\sqrt{15}}{2}$
• A1 $\frac{-3 \pm \sqrt{15}}{2}$