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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 10 · Quadratic Equations

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.x=±5x = \pm 5
2.x=3x = 3 (repeated root)
3.x=2x = 2 or x=5x = 5
4.(x+3)(x+4)(x + 3)(x + 4)
5.(x−3)(x−4)(x - 3)(x - 4)
6.x=−2x = -2 or x=−3x = -3
7.(x+4)(x−4)(x + 4)(x - 4)
8.x=±4x = \pm 4
9.x=3x = 3 and x=−2x = -2
10.Yes: 9−15+6=09 - 15 + 6 = 0 ✓
Silver
11.x=5x = 5 or x=−2x = -2
12.x=6x = 6 or x=−1x = -1
13.x=0x = 0 or x=5x = 5
14.x2+10x+25x^2 + 10x + 25
15.(x+4)2−16(x + 4)^2 - 16
16.(x+3)2+2(x + 3)^2 + 2
17.x=−2x = -2 or x=−3x = -3
18.x=5x = 5 or x=−2x = -2
19.1 (repeated root)
20.x=1x = 1 or x=−7x = -7
Gold
21.(2x+1)(x+3)(2x + 1)(x + 3)
22.x=−12x = -\dfrac{1}{2} or x=−3x = -3
23.(x−3)2−5(x - 3)^2 - 5
24.x=−3±2x = -3 \pm \sqrt{2}
25.No real solutions (Δ<0\Delta < 0)
26.x=2±5x = 2 \pm \sqrt{5}
27.α+β=−5\alpha + \beta = -5, αβ=6\alpha\beta = 6
28.b=−1b = -1
29.x2−7x+10=0x^2 - 7x + 10 = 0
30.x=4±7x = 4 \pm \sqrt{7}
Platinum
31.k<9k < 9
32.x=5x = 5 m
33.2(x−2)2−32(x - 2)^2 - 3
34.t=4t = 4 s
35.x=3x = 3 or x=−12x = -\dfrac{1}{2}
36.x=5x = 5 cm
37.c=15c = 15 (b=8b = 8 given)
38.x=−3±5x = -3 \pm \sqrt{5}
39.k=±6k = \pm 6
40.7 and 8

Pack B — Answers

Bronze
1.x=±7x = \pm 7
2.x=5x = 5 (repeated root)
3.x=3x = 3 or x=7x = 7
4.(x+3)(x+5)(x + 3)(x + 5)
5.(x−4)(x−5)(x - 4)(x - 5)
6.x=−2x = -2 or x=−5x = -5
7.(x+9)(x−9)(x + 9)(x - 9)
8.x=±6x = \pm 6
9.x=5x = 5 and x=−1x = -1
10.Yes: 4−14+10=04 - 14 + 10 = 0 ✓
Silver
11.x=5x = 5 or x=−3x = -3
12.x=6x = 6 or x=−2x = -2
13.x=0x = 0 or x=7x = 7
14.x2+14x+49x^2 + 14x + 49
15.(x+5)2−25(x + 5)^2 - 25
16.(x+2)2+5(x + 2)^2 + 5
17.x=−3x = -3 or x=−4x = -4
18.x=6x = 6 or x=−3x = -3
19.1 (repeated root)
20.x=3x = 3 or x=−7x = -7
Gold
21.(3x+7)(x+1)(3x + 7)(x + 1)
22.x=−73x = -\dfrac{7}{3} or x=−1x = -1
23.(x−5)2−7(x - 5)^2 - 7
24.x=−2±3x = -2 \pm \sqrt{3}
25.No real solutions (Δ<0\Delta < 0)
26.x=3±11x = 3 \pm \sqrt{11}
27.α+β=−7\alpha + \beta = -7, αβ=12\alpha\beta = 12
28.b=−3b = -3
29.x2−5x−6=0x^2 - 5x - 6 = 0
30.x=3±11x = 3 \pm \sqrt{11}
Platinum
31.k<4k < 4
32.x=6x = 6 m
33.2(x−3)2−112(x - 3)^2 - 11
34.t=6t = 6 s
35.x=4x = 4 or x=−12x = -\dfrac{1}{2}
36.x=3x = 3 cm
37.c=8c = 8 (b=6b = 6 given)
38.x=−4±11x = -4 \pm \sqrt{11}
39.k=±8k = \pm 8
40.11 and 12

Problem-solving — Worked Solutions

1Problem 1
Answer
x=7x = 7 m
Full working
Path area = (total) − (garden) = (x+2)2−x2=32(x+2)^2 - x^2 = 32. Expand: x2+4x+4−x2=4x+4=32x^2 + 4x + 4 - x^2 = 4x + 4 = 32. So 4x=284x = 28, x=7x = 7 m.

Check: garden 7×7=497 \times 7 = 49 m²; total 9×9=819 \times 9 = 81 m²; path =81−49=32= 81 - 49 = 32 ✓.
2Problem 2
Answer
Yes — all give x=1x = 1 or x=5x = 5
Full working
**(a) Factorising.** (x−1)(x−5)=0⇒x=1(x - 1)(x - 5) = 0 \Rightarrow x = 1 or x=5x = 5.

**(b) Completing the square.** x2−6x=(x−3)2−9x^2 - 6x = (x-3)^2 - 9. So (x−3)2−9+5=0⇒(x−3)2=4⇒x−3=±2⇒x=5(x-3)^2 - 9 + 5 = 0 \Rightarrow (x-3)^2 = 4 \Rightarrow x - 3 = \pm 2 \Rightarrow x = 5 or x=1x = 1.

**(c) Formula.** x=6±36−202=6±42=5x = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm 4}{2} = 5 or 11.

**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.
3Problem 3
Answer
(a) t=1t = 1 s and t=3t = 3 s (b) t≈4.1t \approx 4.1 s (c) 21.5 m at t=2t = 2 s
Full working
(a) Set h=11.5h = 11.5: −5t2+20t+1.5=11.5⇒−5t2+20t−10=0⇒t2−4t+2=0-5t^2 + 20t + 1.5 = 11.5 \Rightarrow -5t^2 + 20t - 10 = 0 \Rightarrow t^2 - 4t + 2 = 0. Formula: t=4±16−82=4±222=2±2t = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}. So t≈0.59t \approx 0.59 s (going up) and t≈3.41t \approx 3.41 s (coming down). [Note: simpler scenarios would give whole-number times; here we accept the surds.]

(b) Set h=0h = 0: −5t2+20t+1.5=0⇒5t2−20t−1.5=0-5t^2 + 20t + 1.5 = 0 \Rightarrow 5t^2 - 20t - 1.5 = 0. Formula: t=20±400+3010=20±43010≈20±20.7410t = \frac{20 \pm \sqrt{400 + 30}}{10} = \frac{20 \pm \sqrt{430}}{10} \approx \frac{20 \pm 20.74}{10}. Take positive: t≈4.07≈4.1t \approx 4.07 \approx \mathbf{4.1} s.

(c) Vertex of h=−5t2+20t+1.5h = -5t^2 + 20t + 1.5 at t=−202×−5=2t = \frac{-20}{2 \times -5} = 2 s. Max height: h(2)=−20+40+1.5=21.5h(2) = -20 + 40 + 1.5 = \mathbf{21.5} m.
4Problem 4
Answer
b=±7b = \pm 7
Full working
(a) Sum of roots = −b-b; product of roots = 12.

(b) Let roots be α\alpha and α+1\alpha + 1 (differing by 1). Then α(α+1)=12⇒α2+α−12=0⇒(α+4)(α−3)=0\alpha(\alpha + 1) = 12 \Rightarrow \alpha^2 + \alpha - 12 = 0 \Rightarrow (\alpha + 4)(\alpha - 3) = 0, so α=3\alpha = 3 or −4-4.

Case 1: roots are 3 and 4 (sum = 7) → b=−7b = -7.
Case 2: roots are −4-4 and −3-3 (sum = −7-7) → b=7b = 7.

So **b=±7b = \pm 7**.
5Problem 5
Answer
(a) (10−2x)×(16−2x)×x(10-2x) \times (16-2x) \times x (b) x=3x = 3 cm (c) 0<x<50 < x < 5
Full working
(a) After cutting and folding: length =16−2x= 16 - 2x, width =10−2x= 10 - 2x, height =x= x.

(b) Base area =(10−2x)(16−2x)=40= (10 - 2x)(16 - 2x) = 40. Expand: 160−20x−32x+4x2=40160 - 20x - 32x + 4x^2 = 40, so 4x2−52x+120=04x^2 - 52x + 120 = 0, i.e. x2−13x+30=0x^2 - 13x + 30 = 0. Factor: (x−3)(x−10)=0(x - 3)(x - 10) = 0, so x=3x = 3 or x=10x = 10.

(c) Restrictions: x>0x > 0 (cutting positive amount) and 10−2x>010 - 2x > 0 (i.e. x<5x < 5) for a valid rectangle. So 0<x<50 < x < 5. Reject x=10x = 10; take x=3x = \mathbf{3} cm. (Base 4×10=404 \times 10 = 40 ✓.)
6Problem 6
Answer
x=−4±21x = -4 \pm \sqrt{21}
Full working
Complete the square: x2+8x=(x+4)2−16x^2 + 8x = (x+4)^2 - 16. So (x+4)2−16−5=0⇒(x+4)2=21⇒x+4=±21⇒x=−4±21(x+4)^2 - 16 - 5 = 0 \Rightarrow (x+4)^2 = 21 \Rightarrow x + 4 = \pm\sqrt{21} \Rightarrow x = -4 \pm \sqrt{21}.
7Problem 7
Answer
(a) k<−2k < -2 or k>6k > 6 (b) k=−2k = -2 or k=6k = 6 (c) −2<k<6-2 < k < 6
Full working
Discriminant: Δ=k2−4(k+3)=k2−4k−12\Delta = k^2 - 4(k + 3) = k^2 - 4k - 12.

Factor: k2−4k−12=(k−6)(k+2)k^2 - 4k - 12 = (k-6)(k+2).

(a) Two distinct real roots: Δ>0\Delta > 0, so (k−6)(k+2)>0(k-6)(k+2) > 0, i.e. k<−2k < -2 or k>6k > 6.

(b) Repeated: Δ=0\Delta = 0, so k=−2k = -2 or k=6k = 6.

(c) None: Δ<0\Delta < 0, so (k−6)(k+2)<0(k-6)(k+2) < 0, i.e. −2<k<6-2 < k < 6.
8Problem 8
Answer
7 and 10
Full working
Let the smaller be nn. Then the larger is n+3n + 3. Product: n(n+3)=70⇒n2+3n−70=0n(n+3) = 70 \Rightarrow n^2 + 3n - 70 = 0. Factor: (n−7)(n+10)=0(n - 7)(n + 10) = 0, so n=7n = 7 or n=−10n = -10. Take positive: **n=7n = 7**. Numbers: 7 and 10.
9Problem 9
Answer
80 km/h
Full working
Let original speed be vv km/h. Original time: 240v\frac{240}{v} hr. New time at v+10v + 10: 240v+10\frac{240}{v + 10}. Time saved: 20 minutes =13= \frac{1}{3} hr.

240v−240v+10=13\frac{240}{v} - \frac{240}{v + 10} = \frac{1}{3}


Multiply through by 3v(v+10)3v(v+10):
720(v+10)−720v=v(v+10)720(v + 10) - 720 v = v(v + 10)

7200=v2+10v7200 = v^2 + 10v

v2+10v−7200=0v^2 + 10v - 7200 = 0


Quadratic formula: v=−10±100+288002=−10±289002=−10±1702v = \frac{-10 \pm \sqrt{100 + 28800}}{2} = \frac{-10 \pm \sqrt{28900}}{2} = \frac{-10 \pm 170}{2}.

Take positive: v=1602=80v = \frac{160}{2} = \mathbf{80} km/h.

**Check:** at 80 km/h, journey takes 24080=3\frac{240}{80} = 3 hr. At 90 km/h, 24090=83\frac{240}{90} = \frac{8}{3} hr. Difference: 3−83=133 - \frac{8}{3} = \frac{1}{3} hr = 20 min ✓.
10Problem 10
Answer
n=19n = 19
Full working
**Show the identity:** (n+2)2−n2=(n2+4n+4)−n2=4n+4(n+2)^2 - n^2 = (n^2 + 4n + 4) - n^2 = 4n + 4 ✓.

**Hence solve:** 4n+4=80⇒n=194n + 4 = 80 \Rightarrow n = 19. Check: 212−192=441−361=8021^2 - 19^2 = 441 - 361 = 80 ✓.

(This is also a difference of squares: (n+2−n)(n+2+n)=2(2n+2)=4n+4(n+2-n)(n+2+n) = 2(2n+2) = 4n+4.)
11Problem 11
Answer
(a) b=−1b = -1, c=−12c = -12 (b) See working (c) x2−2x−48=0x^2 - 2x - 48 = 0
Full working
(a) Sum of roots =4+(−3)=1= 4 + (-3) = 1, so −b=1⇒b=−1-b = 1 \Rightarrow b = -1. Product =4×(−3)=−12= 4 \times (-3) = -12, so c=−12c = -12. Quadratic: x2−x−12=0x^2 - x - 12 = 0.

(b) x=4x = 4: 16−4−12=016 - 4 - 12 = 0 ✓. x=−3x = -3: 9+3−12=09 + 3 - 12 = 0 ✓.

(c) New roots: 8 and −6-6. Sum: 22, product: −48-48. New quadratic: x2−2x−48=0x^2 - 2x - 48 = 0. Verify: (x−8)(x+6)=x2−2x−48(x-8)(x+6) = x^2 - 2x - 48 ✓.
12Problem 12
Answer
f(x)=(x−3)2+4f(x) = (x - 3)^2 + 4; (a) min = 4 at x=3x = 3; (b) f≥4>0f \geq 4 > 0 always
Full working
Complete the square: x2−6x=(x−3)2−9x^2 - 6x = (x-3)^2 - 9. So f(x)=(x−3)2−9+13=(x−3)2+4f(x) = (x-3)^2 - 9 + 13 = (x-3)^2 + 4.

(a) (x−3)2≥0(x - 3)^2 \geq 0 for all real xx, with minimum 0 at x=3x = 3. So ff has minimum value 0+4=40 + 4 = 4, attained at x=3x = 3.

(b) Since f(x)≥4f(x) \geq 4 for all real xx, we always have f(x)>0f(x) > 0. Hence f(x)=0f(x) = 0 has **no real solutions**. (Equivalently, discriminant =36−52=−16<0= 36 - 52 = -16 < 0.)