Sec 2 Maths: Solving Quadratic Equations by Factorisation, practice questions & worked solutions

(a)

Bring every term to one side rather than dividing by $x$, which would lose the root $x=0$:

$$\begin{aligned}x^2-x&=0\\ x(x-1)&=0\end{aligned}$$
$$\begin{aligned}x=0\quad&\text{or}\quad x=1\end{aligned}$$

(b)

Note: the paper prints the right-hand side as $(3y+2)$, but the word “Hence” and the school answers $y=\dfrac23$ or $y=1$ fit $(3y-2)^2=(3y-2)$, which is solved here.

This is the equation in (a) with $x=3y-2$, so $3y-2=0$ or $3y-2=1$:

$$\begin{aligned}3y-2=0\quad&\text{or}\quad 3y-2=1\\ 3y=2\quad&\text{or}\quad 3y=3\\ y=\frac23\quad&\text{or}\quad y=1\end{aligned}$$

(a)

Substitute $x=\dfrac13$:

$$\begin{aligned}k\left(\frac13\right)^2+(k+2)\left(\frac13\right)-2&=0\\ \frac{k}{9}+\frac{k}{3}+\frac23-2&=0\\ \frac{4k}{9}&=\frac43\\ k&=3\end{aligned}$$

(b)

$$\begin{aligned}3x^2+5x-2&=0\\ (3x-1)(x+2)&=0\end{aligned}$$
$$\begin{aligned}x=\frac13\quad&\text{or}\quad x=-2\end{aligned}$$

The other solution is $x=-2$.

(a)

Substitute $x=3$:

$$\begin{aligned}2(3)^2-b(3)-3&=0\\ 18-3b-3&=0\\ 3b&=15\\ b&=5\end{aligned}$$

(b)

$$\begin{aligned}2x^2-5x-3&=0\\ (2x+1)(x-3)&=0\end{aligned}$$
$$\begin{aligned}x=-\frac12\quad&\text{or}\quad x=3\end{aligned}$$

The other solution is $x=-\dfrac12$.

Multiply both sides by $(x+1)$, where $x\neq-1$:

$$\begin{aligned}20&=(2x+5)(x+1)\\ 20&=2x^2+7x+5\\ 2x^2+7x-15&=0\\ (2x-3)(x+5)&=0\end{aligned}$$
$$\begin{aligned}x=1.5\quad&\text{or}\quad x=-5\end{aligned}$$

(a)

The right angle is at $Q$, so $PR$ is the hypotenuse. By Pythagoras’ theorem:

$$\begin{aligned}(2x+5)^2&=(x+3)^2+(x+4)^2\\ 4x^2+20x+25&=x^2+6x+9+x^2+8x+16\\ 4x^2+20x+25&=2x^2+14x+25\\ 2x^2+6x&=0\quad\text{(shown)}\end{aligned}$$

(b)

$$\begin{aligned}2x(x+3)&=0\end{aligned}$$
$$\begin{aligned}x=0\quad&\text{or}\quad x=-3\end{aligned}$$

(c)

If $x=-3$, then $PQ=-3+3=0$ cm and $PR=2(-3)+5=-1$ cm. A side of a triangle must have positive length, so $x=-3$ is rejected.

(d)

With $x=0$: $PQ=3$ cm and $QR=4$ cm.

$$\begin{aligned}\text{Area}&=\frac12\times3\times4\\ &=6\text{ cm}^2\end{aligned}$$

(a)

Area of a trapezium $=\dfrac12\times(\text{sum of parallel sides})\times\text{height}$. The parallel sides are $AB$ and $DC$, and the height is $AD$:

$$\begin{aligned}\frac12\left[(x-1)+(3x-1)\right](x+3)&=15\\ \frac12(4x-2)(x+3)&=15\\ (2x-1)(x+3)&=15\\ 2x^2+6x-x-3&=15\\ 2x^2+5x-18&=0\quad\text{(shown)}\end{aligned}$$

(b)

$$\begin{aligned}(2x+9)(x-2)&=0\end{aligned}$$
$$\begin{aligned}x=-\frac92\quad&\text{or}\quad x=2\end{aligned}$$

(c)

If $x=-\dfrac92$, then $AD=-\dfrac92+3=-\dfrac32$ cm, and a length cannot be negative. So $x=-\dfrac92$ is rejected and $x=2$.

(a)

Area of a parallelogram $=\text{base}\times\text{perpendicular height}$:

$$\begin{aligned}(18-x)(x+1)&=78\\ 18x+18-x^2-x&=78\\ -x^2+17x-60&=0\\ x^2-17x+60&=0\quad\text{(shown)}\end{aligned}$$

(b)

$$\begin{aligned}(x-5)(x-12)&=0\end{aligned}$$
$$\begin{aligned}x=5\quad&\text{or}\quad x=12\end{aligned}$$

Since $x$ is less than $10$, $x=12$ is rejected, so $x=5$.

Each row has $5$ numbers, so moving one square right adds $1$ and moving one square down adds $5$. In the shape, the top-right number is $1$ more than the top-left, and the bottom number is $5$ more than the top-right.

(a)

Top-left $=74-1=73$ and bottom $=74+5=79$.

(b)(i)

The three numbers are $x$, $x+1$ and $x+6$, so the largest is $x+6$.

(b)(ii)

$$\begin{aligned}x(x+6)&=135\\ x^2+6x&=135\\ x^2+6x-135&=0\quad\text{(shown)}\end{aligned}$$

(b)(iii)

$$\begin{aligned}(x+15)(x-9)&=0\end{aligned}$$
$$\begin{aligned}x=-15\quad&\text{or}\quad x=9\end{aligned}$$

(b)(iv)

The numbers on the grid are positive, so $x=9$. The three numbers are $9$, $10$ and $15$:

$$\begin{aligned}9+10+15&=34\end{aligned}$$

(a)

$$\begin{aligned}\text{James' speed}&=\frac{300}{x}\text{ km/h}\end{aligned}$$

(b)

$$\begin{aligned}\text{Ben's speed}&=\frac{300}{x+2}\text{ km/h}\end{aligned}$$

(c)

James’ speed minus the train’s speed is $40$ km/h:

$$\begin{aligned}\frac{300}{x}-\frac{300}{x+2}&=40\\ 300(x+2)-300x&=40x(x+2)\\ 600&=40x^2+80x\\ 40x^2+80x-600&=0\\ x^2+2x-15&=0\quad\text{(shown)}\end{aligned}$$

(d)

$$\begin{aligned}(x+5)(x-3)&=0\end{aligned}$$
$$\begin{aligned}x=-5\quad&\text{or}\quad x=3\end{aligned}$$

(e)

Time cannot be negative, so $x=3$. The train takes $3+2=5$ hours:

$$\begin{aligned}\text{Train speed}&=\frac{300}{5}\\ &=60\text{ km/h}\end{aligned}$$

Distance $=$ speed $\times$ time.

(a)

$$\begin{aligned}\text{Walking distance}&=x(x-1)\text{ km}\end{aligned}$$

(b)

$$\begin{aligned}\text{Cycling distance}&=(2x+1)(x+2)\text{ km}\end{aligned}$$

(c)

$$\begin{aligned}x(x-1)+(2x+1)(x+2)&=97\\ x^2-x+2x^2+4x+x+2&=97\\ 3x^2+4x+2&=97\\ 3x^2+4x-95&=0\quad\text{(shown)}\end{aligned}$$

(d)

$$\begin{aligned}(x-5)(3x+19)&=0\end{aligned}$$
$$\begin{aligned}x=5\quad&\text{or}\quad x=-\frac{19}{3}\end{aligned}$$

(e)

$x$ is the walking time in hours, and a time cannot be negative. So $x=-\dfrac{19}{3}$ is rejected.

(f)

Total time $=x+(x+2)=2x+2$ hours. With $x=5$:

$$\begin{aligned}2(5)+2&=12\text{ hours}\end{aligned}$$