Solutions — 5G Worksheet
Year 10 Mathematics Core · Cambridge Ch 5, §5G · Mr Wong
ANSWER KEY
Warm-up
Q1. Already factorised — set each factor to $0$:
a $x=0$ or $x=6$
b $x=5$ or $x=-1$
c $x=\tfrac12$ or $x=-4$
d $x=-\tfrac23$ or $x=\tfrac34$
Part A — Common factor & difference of two squares
Q2.
a $x=0$ or $x=-5$ $x(x+5)=0$
b $x=7$ or $x=-7$ $(x-7)(x+7)=0$
c $x=5$ or $x=-5$ $3(x^2-25)=0$
d $x=\tfrac53$ or $x=-\tfrac53$ $(3x-5)(3x+5)=0$
Part B — Monic trinomials
Q3.
a $x=-5$ or $x=-2$ $(x+5)(x+2)=0$
b $x=4$ or $x=5$ $(x-4)(x-5)=0$
c $x=-5$ or $x=3$ $(x+5)(x-3)=0$
d $x=-3$ (one solution) $(x+3)^2=0$
Part C — Non-monic trinomials
Q4.
a $x=\tfrac12$ or $x=-3$ $2x^2+6x-x-3=0 \Rightarrow (2x-1)(x+3)=0$
b $x=\tfrac43$ or $x=2$ $3x^2-4x-6x+8=0 \Rightarrow (3x-4)(x-2)=0$
Part D — Disguised quadratics
Q5.
a $x=6$ or $x=-5$ $x^2-x-30=0 \Rightarrow (x-6)(x+5)=0$
b $x=10$ or $x=-6$ $x^2-4x-60=0 \Rightarrow (x-10)(x+6)=0$
Challenge
Q6. $\dfrac{x+12}{x}=x$:
· $x+12=x^2 \Rightarrow x^2-x-12=0 \Rightarrow (x-4)(x+3)=0$, so
$x=4$ or $x=-3$ — two solutions.