Algebra — Introduction & Intermediate

Question 1 (10)

(a) Area $=(2x+3)(x-1)$. Expand: $2x^2 - 2x + 3x - 3 = 2x^2 + x - 3$. (1 setup, 2 expansion)

(b) Inner region has length reduced by $2$ (1 m each end) and width reduced by $2$: inner length $=(2x+3)-2 = 2x+1$; inner width $=(x-1)-2 = x-3$. (2) Inner area $=(2x+1)(x-3) = 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3$. (2)

(c) $2x^2 + x - 3 = 65 \Rightarrow 2x^2 + x - 68 = 0$. (1) Factor/quadratic formula: discriminant $=1 + 8\cdot68 = 545$... check: use formula $x = \dfrac{-1 \pm \sqrt{1 + 544}}{4} = \dfrac{-1 \pm \sqrt{545}}{4}$.

(Correction for clean roots — solve directly): $2x^2 + x - 68 = 0$ factors as $(2x + 17)(x - 4)=0$ since $(2x+17)(x-4)=2x^2 -8x +17x -68 = 2x^2+9x-68$ ✗.

Use quadratic formula honestly: $x=\dfrac{-1+\sqrt{545}}{4}\approx 5.59$ or $x\approx-6.09$. Reject the negative root (width $x-1$ must be positive, so $x>1$). (2) Valid solution $x = \dfrac{-1+\sqrt{545}}{4}$. Reject negative as it gives negative dimensions. (1)

Question 2 (10)

(a) By Factor Theorem, $P(2)=0$: $8 + 4a - 26 + b = 0 \Rightarrow 4a + b = 18.$ (2) By Remainder Theorem, $P(-3) = -30$: $-27 + 9a + 39 + b = -30 \Rightarrow 9a + b = -42.$ (2) Subtract: $5a = -60 \Rightarrow a = -12$, then $b = 18 - 4(-12) = 66$. (2)

(b) $P(x) = x^3 - 12x^2 - 13x + 66$. $(x-2)$ is a factor. Divide: $x^3 - 12x^2 -13x + 66 \div (x-2) = x^2 - 10x - 33$. (2) Factor $x^2 - 10x - 33 = (x - 11)(x + 3)$ since $-11\cdot3=-33$, $-11+3=-8$ ✗. Check: need product $-33$, sum $-10$: roots $-11,\,3$? $-11+3=-8$. Not matching. Use formula: $x = \dfrac{10 \pm \sqrt{100+132}}{2} = \dfrac{10\pm\sqrt{232}}{2}$ — not integer.

(Recompute quotient): proper synthetic division of $[1,-12,-13,66]$ by root $2$: bring 1; $1\cdot2=2$, $-12+2=-10$; $-10\cdot2=-20$, $-13-20=-33$; $-33\cdot2=-66$, $66-66=0$ ✓. So quotient $x^2-10x-33$. This does not factor over integers; complete factorisation: $P(x) = (x-2)(x^2 - 10x - 33)$, remaining roots $5\pm\sqrt{58}$. (2)

Question 3 (10)

(a) $\sqrt{2x+7} = x+2$. Square: $2x+7 = x^2 + 4x + 4$. (2) $0 = x^2 + 2x - 3 = (x+3)(x-1)$, so $x=-3$ or $x=1$. (2) Check: $x=1$: $\sqrt9-1 = 3-1=2$ ✓. $x=-3$: $\sqrt1-(-3)=1+3=4\ne2$ ✗ (extraneous). Solution: $x=1$. (1)

(b) Rationalise each: $\dfrac{3}{\sqrt5-\sqrt2}=\dfrac{3(\sqrt5+\sqrt2)}{5-2}=\sqrt5+\sqrt2.$ (2) $\dfrac{2}{\sqrt5+\sqrt2}=\dfrac{2(\sqrt5-\sqrt2)}{3}=\dfrac{2\sqrt5-2\sqrt2}{3}.$ (2) Difference: $\sqrt5+\sqrt2 - \dfrac{2\sqrt5 - 2\sqrt2}{3} = \dfrac{3\sqrt5+3\sqrt2 - 2\sqrt5 + 2\sqrt2}{3} = \dfrac{\sqrt5 + 5\sqrt2}{3}.$ So $p=\tfrac13,\ q=\tfrac53$. (1)

Question 4 (10)

(a) Let numbers be $x$ and $x-4$ (both positive). $x^2 + (x-4)^2 = 40 \Rightarrow 2x^2 - 8x + 16 = 40 \Rightarrow 2x^2 - 8x - 24 = 0$ $\Rightarrow x^2 - 4x - 12 = 0 \Rightarrow (x-6)(x+2)=0.$ (3) $x=6$ (reject $-2$). Numbers: $6$ and $2$. (2)

(b) Reciprocals $\tfrac16,\ \tfrac12$. Sum $=\tfrac16+\tfrac12 = \tfrac{2}{3}$; product $=\tfrac{1}{12}$. (2) Equation $x^2 - (\text{sum})x + (\text{product}) = 0$: $x^2 - \tfrac23 x + \tfrac1{12} = 0 \Rightarrow 12x^2 - 8x + 1 = 0.$ (3)

Question 5 (10)

(a) $|2x-5|<3 \Rightarrow -3 < 2x-5 < 3 \Rightarrow 2 < 2x < 8 \Rightarrow 1 < x < 4.$ (3) Number line: open circles at $1$ and $4$, shaded between. (1)

(b) $3x-1\le8 \Rightarrow x\le3$; $\tfrac x2+1>0 \Rightarrow x>-2$. (3) AND: $-2 < x \le 3$. (1)

(c) Equal roots $\Rightarrow$ discriminant $=0$: $k^2 - 36 = 0 \Rightarrow k=\pm6.$ (2)

[
  {"claim":"Q2 a,b values give factor theorem & remainder conditions",
   "code":"a,b,x=symbols('a b x'); sol=solve([4*a+b-18,9*a+b+42],[a,b]); result=(sol[a]==-12 and sol[b]==66)"},
  {"claim":"Q2 division quotient is x^2-10x-33",
   "code":"x=symbols('x'); P=x**3-12*x**2-13*x+66; q=simplify(P/(x-2)); result=(expand(q)==x**2-10*x-33)"},
  {"claim":"Q3a only valid root is x=1",
   "code":"x=symbols('x'); sols=solve(sqrt(2*x+7)-x-2,x); result=(sols==[1])"},
  {"claim":"Q3b simplifies to (sqrt5+5sqrt2)/3",
   "code":"e=3/(sqrt(5)-sqrt(2))-2/(sqrt(5)+sqrt(2)); result=simplify(e-(sqrt(5)+5*sqrt(2))/3)==0"},
  {"claim":"Q4 numbers are 6 and 2",
   "code":"x=symbols('x'); s=solve(x**2-4*x-12,x); result=(6 in s and -2 in s)"},
  {"claim":"Q4b reciprocal quadratic 12x^2-8x+1 has roots 1/6,1/2",
   "code":"x=symbols('x'); r=set(solve(12*x**2-8*x+1,x)); result=(r=={Rational(1,6),Rational(1,2)})"},
  {"claim":"Q5c equal roots when k=+-6",
   "code":"k=symbols('k'); result=set(solve(k**2-36,k))=={6,-6}"}
]