Answer Key

Quadratic Equations

Pack A — Answers

#	Question	Answer
1	Solve $x^2 = 25$.	$x = \pm 5$
2	Solve $(x - 3)^2 = 0$.	$x = 3$ (repeated root)
3	Solve $(x - 2)(x - 5) = 0$.	$x = 2$ or $x = 5$
4	Factorise $x^2 + 7x + 12$.	$(x + 3)(x + 4)$
5	Factorise $x^2 - 7x + 12$.	$(x - 3)(x - 4)$
6	Solve $x^2 + 5x + 6 = 0$.	$x = -2$ or $x = -3$
7	Factorise $x^2 - 16$.	$(x + 4)(x - 4)$
8	Solve $x^2 - 16 = 0$.	$x = \pm 4$
9	State the roots of $y = (x - 3)(x + 2)$.	$x = 3$ and $x = -2$
10	Verify that $x = 3$ is a root of $x^2 - 5x + 6 = 0$.	Yes: $9 - 15 + 6 = 0$ ✓
11	Solve $x^2 - 3x - 10 = 0$.	$x = 5$ or $x = -2$
12	Solve $x^2 = 5x + 6$.	$x = 6$ or $x = -1$
13	Solve $x^2 = 5x$.	$x = 0$ or $x = 5$
14	Expand $(x + 5)^2$.	$x^2 + 10x + 25$
15	Write $x^2 + 8x$ as $(x + p)^2 - q$.	$(x + 4)^2 - 16$
16	Write $x^2 + 6x + 11$ in completed-square form $(x + p)^2 + q$.	$(x + 3)^2 + 2$
17	Use the quadratic formula to solve $x^2 + 5x + 6 = 0$.	$x = -2$ or $x = -3$
18	Solve $(x + 1)(x - 4) = 6$.	$x = 5$ or $x = -2$
19	How many real roots does $x^2 - 4x + 4 = 0$ have?	1 (repeated root)
20	Solve $(x + 3)^2 = 16$.	$x = 1$ or $x = -7$
21	Factorise $2x^2 + 7x + 3$.	$(2x + 1)(x + 3)$
22	Use the quadratic formula to solve $2x^2 + 7x + 3 = 0$.	$x = -\dfrac{1}{2}$ or $x = -3$
23	Write $x^2 - 6x + 4$ in completed-square form.	$(x - 3)^2 - 5$
24	Solve $x^2 + 6x + 7 = 0$ by completing the square (give exact answers).	$x = -3 \pm \sqrt{2}$
25	Solve $x^2 + 2x + 5 = 0$.	No real solutions ($\Delta < 0$)
26	Solve $x^2 - 4x - 1 = 0$, giving exact answers.	$x = 2 \pm \sqrt{5}$
27	The roots of $x^2 + 5x + 6 = 0$ are $\alpha$ and $\beta$. State $\alpha + \beta$ and $\alpha \beta$.	$\alpha + \beta = -5$, $\alpha\beta = 6$
28	If $x = 3$ is a root of $x^2 + bx - 6 = 0$, find $b$.	$b = -1$
29	Find a monic quadratic equation with roots $2$ and $5$.	$x^2 - 7x + 10 = 0$
30	Solve $(x - 4)^2 = 7$, giving exact answers.	$x = 4 \pm \sqrt{7}$
31	For what values of $k$ does $x^2 + 6x + k = 0$ have two distinct real roots?	$k < 9$
32	A rectangular garden is $x$ m wide and $(x + 3)$ m long. Its area is $40$ m². Find $x$.	$x = 5$ m
33	Write $2x^2 - 8x + 5$ in the form $a(x + p)^2 + q$.	$2(x - 2)^2 - 3$
34	A ball is thrown upward and its height $h$ (m) after $t$ s is given by $h = 20t - 5t^2$. When does it hit the ground (other than at $t = 0$)?	$t = 4$ s
35	Solve $2x^2 - 5x - 3 = 0$ exactly.	$x = 3$ or $x = -\dfrac{1}{2}$
36	A right-angled triangle has legs of length $x$ and $(x + 7)$ cm. Its hypotenuse is $13$ cm. Find $x$.	$x = 5$ cm
37	The quadratic $x^2 - 8x + c = 0$ has roots that differ by 2. If one root is $5$, find $c$ and $b$.	$c = 15$ ($b = 8$ given)
38	Solve $x^2 + 6x + 4 = 0$ exactly, in surd form.	$x = -3 \pm \sqrt{5}$
39	The equation $x^2 + kx + 9 = 0$ has exactly one (repeated) real root. Find $k$.	$k = \pm 6$
40	The product of two consecutive positive integers is $56$. Find the integers.	7 and 8

Pack B — Answers

#	Question	Answer
1	Solve $x^2 = 49$.	$x = \pm 7$
2	Solve $(x - 5)^2 = 0$.	$x = 5$ (repeated root)
3	Solve $(x - 3)(x - 7) = 0$.	$x = 3$ or $x = 7$
4	Factorise $x^2 + 8x + 15$.	$(x + 3)(x + 5)$
5	Factorise $x^2 - 9x + 20$.	$(x - 4)(x - 5)$
6	Solve $x^2 + 7x + 10 = 0$.	$x = -2$ or $x = -5$
7	Factorise $x^2 - 81$.	$(x + 9)(x - 9)$
8	Solve $x^2 - 36 = 0$.	$x = \pm 6$
9	State the roots of $y = (x - 5)(x + 1)$.	$x = 5$ and $x = -1$
10	Verify that $x = 2$ is a root of $x^2 - 7x + 10 = 0$.	Yes: $4 - 14 + 10 = 0$ ✓
11	Solve $x^2 - 2x - 15 = 0$.	$x = 5$ or $x = -3$
12	Solve $x^2 = 4x + 12$.	$x = 6$ or $x = -2$
13	Solve $x^2 = 7x$.	$x = 0$ or $x = 7$
14	Expand $(x + 7)^2$.	$x^2 + 14x + 49$
15	Write $x^2 + 10x$ as $(x + p)^2 - q$.	$(x + 5)^2 - 25$
16	Write $x^2 + 4x + 9$ in completed-square form $(x + p)^2 + q$.	$(x + 2)^2 + 5$
17	Use the quadratic formula to solve $x^2 + 7x + 12 = 0$.	$x = -3$ or $x = -4$
18	Solve $(x + 2)(x - 5) = 8$.	$x = 6$ or $x = -3$
19	How many real roots does $x^2 - 6x + 9 = 0$ have?	1 (repeated root)
20	Solve $(x + 2)^2 = 25$.	$x = 3$ or $x = -7$
21	Factorise $3x^2 + 10x + 7$.	$(3x + 7)(x + 1)$
22	Use the quadratic formula to solve $3x^2 + 10x + 7 = 0$.	$x = -\dfrac{7}{3}$ or $x = -1$
23	Write $x^2 - 10x + 18$ in completed-square form.	$(x - 5)^2 - 7$
24	Solve $x^2 + 4x + 1 = 0$ by completing the square (give exact answers).	$x = -2 \pm \sqrt{3}$
25	Solve $x^2 + 4x + 10 = 0$.	No real solutions ($\Delta < 0$)
26	Solve $x^2 - 6x - 2 = 0$, giving exact answers.	$x = 3 \pm \sqrt{11}$
27	The roots of $x^2 + 7x + 12 = 0$ are $\alpha$ and $\beta$. State $\alpha + \beta$ and $\alpha \beta$.	$\alpha + \beta = -7$, $\alpha\beta = 12$
28	If $x = 4$ is a root of $x^2 + bx - 4 = 0$, find $b$.	$b = -3$
29	Find a monic quadratic equation with roots $-1$ and $6$.	$x^2 - 5x - 6 = 0$
30	Solve $(x - 3)^2 = 11$, giving exact answers.	$x = 3 \pm \sqrt{11}$
31	For what values of $k$ does $x^2 + 4x + k = 0$ have two distinct real roots?	$k < 4$
32	A rectangular garden is $x$ m wide and $(x + 4)$ m long. Its area is $60$ m². Find $x$.	$x = 6$ m
33	Write $2x^2 - 12x + 7$ in the form $a(x + p)^2 + q$.	$2(x - 3)^2 - 11$
34	A ball is thrown upward and its height $h$ (m) after $t$ s is given by $h = 30t - 5t^2$. When does it hit the ground (other than at $t = 0$)?	$t = 6$ s
35	Solve $2x^2 - 7x - 4 = 0$ exactly.	$x = 4$ or $x = -\dfrac{1}{2}$
36	A right-angled triangle has legs of length $x$ and $(x + 1)$ cm. Its hypotenuse is $5$ cm. Find $x$.	$x = 3$ cm
37	The quadratic $x^2 - 6x + c = 0$ has roots that differ by 2. If one root is $4$, find $c$ and $b$.	$c = 8$ ($b = 6$ given)
38	Solve $x^2 + 6x + 4 = 0$ exactly, in surd form.	$x = -4 \pm \sqrt{11}$
39	The equation $x^2 + kx + 9 = 0$ has exactly one (repeated) real root. Find $k$.	$k = \pm 8$
40	The product of two consecutive positive integers is $132$. Find the integers.	11 and 12

Problems — Worked Solutions

**Garden path.** A square garden has side $x$ m. A path of uniform width $1$ m is laid around the outside. The total area of garden + path is $(x + 2)^2$ m². If the area of the path alone is $32$ m², find $x$.

Answer

$x = 7$ m

Path area = (total) − (garden) = $(x+2)^2 - x^2 = 32$. Expand: $x^2 + 4x + 4 - x^2 = 4x + 4 = 32$. So $4x = 28$, $x = 7$ m. Check: garden $7 \times 7 = 49$ m²; total $9 \times 9 = 81$ m²; path $= 81 - 49 = 32$ ✓.

**Three methods.** Solve $x^2 - 6x + 5 = 0$ three ways: (a) By factorising. (b) By completing the square. (c) Using the quadratic formula. Do all three methods give the same answers?

Answer

Yes — all give $x = 1$ or $x = 5$

**(a) Factorising.** $(x - 1)(x - 5) = 0 \Rightarrow x = 1$ or $x = 5$. **(b) Completing the square.** $x^2 - 6x = (x-3)^2 - 9$. So $(x-3)^2 - 9 + 5 = 0 \Rightarrow (x-3)^2 = 4 \Rightarrow x - 3 = \pm 2 \Rightarrow x = 5$ or $x = 1$. **(c) Formula.** $x = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm 4}{2} = 5$ or $1$. **All three methods give the same roots.** Choice of method is a matter of efficiency: factorising is quickest *when* the factors are easy to spot; completing the square works for any quadratic; the formula always works.

**Projectile.** A ball is thrown vertically upward from height $1.5$ m at $20$ m/s. Its height after $t$ seconds is $$h(t) = -5t^2 + 20t + 1.5.$$ (a) When is the ball at height $11.5$ m? (b) When does it hit the ground (1 d.p.)? (c) What is the maximum height reached?

Answer

(a) $t = 1$ s and $t = 3$ s (b) $t \approx 4.1$ s (c) 21.5 m at $t = 2$ s

(a) Set $h = 11.5$: $-5t^2 + 20t + 1.5 = 11.5 \Rightarrow -5t^2 + 20t - 10 = 0 \Rightarrow t^2 - 4t + 2 = 0$. Formula: $t = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}$. So $t \approx 0.59$ s (going up) and $t \approx 3.41$ s (coming down). [Note: simpler scenarios would give whole-number times; here we accept the surds.] (b) Set $h = 0$: $-5t^2 + 20t + 1.5 = 0 \Rightarrow 5t^2 - 20t - 1.5 = 0$. Formula: $t = \frac{20 \pm \sqrt{400 + 30}}{10} = \frac{20 \pm \sqrt{430}}{10} \approx \frac{20 \pm 20.74}{10}$. Take positive: $t \approx 4.07 \approx \mathbf{4.1}$ s. (c) Vertex of $h = -5t^2 + 20t + 1.5$ at $t = \frac{-20}{2 \times -5} = 2$ s. Max height: $h(2) = -20 + 40 + 1.5 = \mathbf{21.5}$ m.

**Find unknown coefficient.** The equation $x^2 + bx + 12 = 0$ has roots that differ by 1. (a) Write expressions for the sum and product of the roots in terms of $b$. (b) Use part (a) and the given condition to find $b$.

Answer

$b = \pm 7$

(a) Sum of roots = $-b$; product of roots = 12. (b) Let roots be $\alpha$ and $\alpha + 1$ (differing by 1). Then $\alpha(\alpha + 1) = 12 \Rightarrow \alpha^2 + \alpha - 12 = 0 \Rightarrow (\alpha + 4)(\alpha - 3) = 0$, so $\alpha = 3$ or $-4$. Case 1: roots are 3 and 4 (sum = 7) → $b = -7$. Case 2: roots are $-4$ and $-3$ (sum = $-7$) → $b = 7$. So **$b = \pm 7$**.

**Box without a lid.** A rectangular piece of card is 10 cm wide and 16 cm long. Equal squares of side $x$ cm are cut from each corner, and the sides folded up to make an open box. (a) Find expressions for the dimensions of the box. (b) Find the value of $x$ that gives a base area of $40$ cm². (c) State any restrictions on $x$.

Answer

(a) $(10-2x) \times (16-2x) \times x$ (b) $x = 3$ cm (c) $0 < x < 5$

(a) After cutting and folding: length $= 16 - 2x$, width $= 10 - 2x$, height $= x$. (b) Base area $= (10 - 2x)(16 - 2x) = 40$. Expand: $160 - 20x - 32x + 4x^2 = 40$, so $4x^2 - 52x + 120 = 0$, i.e. $x^2 - 13x + 30 = 0$. Factor: $(x - 3)(x - 10) = 0$, so $x = 3$ or $x = 10$. (c) Restrictions: $x > 0$ (cutting positive amount) and $10 - 2x > 0$ (i.e. $x < 5$) for a valid rectangle. So $0 < x < 5$. Reject $x = 10$; take $x = \mathbf{3}$ cm. (Base $4 \times 10 = 40$ ✓.)

**Solving by completing the square.** Solve $x^2 + 8x - 5 = 0$, giving exact answers.

Answer

$x = -4 \pm \sqrt{21}$

Complete the square: $x^2 + 8x = (x+4)^2 - 16$. So $(x+4)^2 - 16 - 5 = 0 \Rightarrow (x+4)^2 = 21 \Rightarrow x + 4 = \pm\sqrt{21} \Rightarrow x = -4 \pm \sqrt{21}$.

**Discriminant analysis.** The equation $x^2 + kx + (k+3) = 0$ has (a) two distinct real roots — find the values of $k$. (b) one repeated real root — find $k$. (c) no real roots — find $k$.

Answer

(a) $k < -2$ or $k > 6$ (b) $k = -2$ or $k = 6$ (c) $-2 < k < 6$

Discriminant: $\Delta = k^2 - 4(k + 3) = k^2 - 4k - 12$. Factor: $k^2 - 4k - 12 = (k-6)(k+2)$. (a) Two distinct real roots: $\Delta > 0$, so $(k-6)(k+2) > 0$, i.e. $k < -2$ or $k > 6$. (b) Repeated: $\Delta = 0$, so $k = -2$ or $k = 6$. (c) None: $\Delta < 0$, so $(k-6)(k+2) < 0$, i.e. $-2 < k < 6$.

**Two number puzzle.** Two positive numbers differ by 3, and their product is 70. Find them.

Answer

7 and 10

Let the smaller be $n$. Then the larger is $n + 3$. Product: $n(n+3) = 70 \Rightarrow n^2 + 3n - 70 = 0$. Factor: $(n - 7)(n + 10) = 0$, so $n = 7$ or $n = -10$. Take positive: **$n = 7$**. Numbers: 7 and 10.

**Speed problem.** A train travels 240 km. If it had travelled 10 km/h faster, the journey would have taken 20 minutes less. Find the original speed.

Answer

80 km/h

Let original speed be $v$ km/h. Original time: $\frac{240}{v}$ hr. New time at $v + 10$: $\frac{240}{v + 10}$. Time saved: 20 minutes $= \frac{1}{3}$ hr. $$\frac{240}{v} - \frac{240}{v + 10} = \frac{1}{3}$$ Multiply through by $3v(v+10)$: $$720(v + 10) - 720 v = v(v + 10)$$ $$7200 = v^2 + 10v$$ $$v^2 + 10v - 7200 = 0$$ Quadratic formula: $v = \frac{-10 \pm \sqrt{100 + 28800}}{2} = \frac{-10 \pm \sqrt{28900}}{2} = \frac{-10 \pm 170}{2}$. Take positive: $v = \frac{160}{2} = \mathbf{80}$ km/h. **Check:** at 80 km/h, journey takes $\frac{240}{80} = 3$ hr. At 90 km/h, $\frac{240}{90} = \frac{8}{3}$ hr. Difference: $3 - \frac{8}{3} = \frac{1}{3}$ hr = 20 min ✓.

**Number identities.** Show that $(n+2)^2 - n^2 = 4n + 4$ for any integer $n$. Hence, find two positive integers $n$ such that $(n+2)^2 - n^2 = 80$.

Answer

$n = 19$

**Show the identity:** $(n+2)^2 - n^2 = (n^2 + 4n + 4) - n^2 = 4n + 4$ ✓. **Hence solve:** $4n + 4 = 80 \Rightarrow n = 19$. Check: $21^2 - 19^2 = 441 - 361 = 80$ ✓. (This is also a difference of squares: $(n+2-n)(n+2+n) = 2(2n+2) = 4n+4$.)

**Roots and coefficients.** The quadratic $x^2 + bx + c = 0$ has roots $\alpha = 4$ and $\beta = -3$. (a) Use Vieta's formulas to find $b$ and $c$. (b) Verify by substituting each root into the equation. (c) Construct a new quadratic with roots $2\alpha$ and $2\beta$.

Answer

(a) $b = -1$, $c = -12$ (b) See working (c) $x^2 - 2x - 48 = 0$

(a) Sum of roots $= 4 + (-3) = 1$, so $-b = 1 \Rightarrow b = -1$. Product $= 4 \times (-3) = -12$, so $c = -12$. Quadratic: $x^2 - x - 12 = 0$. (b) $x = 4$: $16 - 4 - 12 = 0$ ✓. $x = -3$: $9 + 3 - 12 = 0$ ✓. (c) New roots: 8 and $-6$. Sum: $2$, product: $-48$. New quadratic: $x^2 - 2x - 48 = 0$. Verify: $(x-8)(x+6) = x^2 - 2x - 48$ ✓.

**Vertex form.** Write $f(x) = x^2 - 6x + 13$ in completed-square form and hence (a) state the minimum value of $f$ and where it occurs; (b) explain why $f(x) = 0$ has no real solutions.

Answer

$f(x) = (x - 3)^2 + 4$; (a) min = 4 at $x = 3$; (b) $f \geq 4 > 0$ always

Complete the square: $x^2 - 6x = (x-3)^2 - 9$. So $f(x) = (x-3)^2 - 9 + 13 = (x-3)^2 + 4$. (a) $(x - 3)^2 \geq 0$ for all real $x$, with minimum 0 at $x = 3$. So $f$ has minimum value $0 + 4 = 4$, attained at $x = 3$. (b) Since $f(x) \geq 4$ for all real $x$, we always have $f(x) > 0$. Hence $f(x) = 0$ has **no real solutions**. (Equivalently, discriminant $= 36 - 52 = -16 < 0$.)