Sec 4 E Maths: Quadratic Equations and Completing the Square, practice questions & worked solutions

$$x^2-20x-a$$

$$=(x-10)^2-10^2-a$$

$$=(x+b)^2-125$$

$$\therefore b=-10,\quad-10^2-a=-125$$

$$a=25$$

Answer: $a=25,\ b=-10$

(a)

$$\begin{aligned}x^2-5x+9&=(x-2.5)^2-\frac{25}4+9\\&=\frac{11}4+(x-2.5)^2\end{aligned}$$

$p=\dfrac{11}4$ or $p=2.75$

(b)

$x=\dfrac52$ or $x=2.5$

Answer: (a) $p=\dfrac{11}4$ (b) $x=2.5$

$$\begin{aligned}2(mn)^2-2mn&=3mn+3\\2m^2n^2-5mn-3&=0\\(mn-3)(2mn+1)&=0\\mn=3\quad\text{or}\quad mn&=-\frac12\\n=\frac3m\quad\text{or}\quad n&=-\frac1{2m}\end{aligned}$$

Answer: $n=\dfrac3m$ or $n=-\dfrac1{2m}$

(a)

$$\begin{aligned}-x^2+5x-7&=-(x^2-5x+7)\\&=-\left[\left(x-\dfrac52\right)^2-\left(-\dfrac52\right)^2+7\right]\\&=-\left(x-\dfrac52\right)^2-\dfrac34\end{aligned}$$

(b)

The shape of the curve opens downwards indefinitely and since the maximum point of the curve is $\left(\dfrac52,-\dfrac34\right)$, the curve lies entirely below the $x$-axis. There is no intersection with the $x$-axis and hence no solution.

Method 2

The maximum point is at $\left(\dfrac52,-\dfrac34\right)$, therefore the graph will not cut $x$ axis and hence no solution.

Answer: (a) $-\left(x-\dfrac52\right)^2-\dfrac34$   (b) the maximum value of the expression is $-\dfrac34$, which is below zero, so the curve never meets the $x$ axis

$$\begin{aligned}x^2-12x+25&=0\\(x-6)^2-36+25&=0\\(x-6)^2-11&=0\\(x-6)^2&=11\end{aligned}$$

$$x-6=\sqrt{11}\quad\text{or}\quad x-6=-\sqrt{11}$$

$$x=6+\sqrt{11}\quad\text{or}\quad x=6-\sqrt{11}$$

$$x=9.31662479\quad\text{or}\quad x=2.68337521$$

$$x=9.32\quad\text{or}\quad x=2.68$$

Answer: $x=9.32\quad\text{or}\quad x=2.68$

Completed square form:

$$\begin{aligned}y&=-\left(x+\frac32\right)^2+p\\&=-\left(x^2+3x+\frac94\right)+p\\&=-x^2-3x-\frac94+p\end{aligned}$$

Comparing coefficients with $y=-x^2+mx+3.5$:

$$m=-3;$$

$$-\frac94+p=3.5$$

$$\rightarrow p=\frac{23}{4}$$

Alternative method

Completed square form:

$$y=-\left(x-\frac m2\right)^2+\frac{m^2}{4}+3.5$$

Comparing coefficients with $y=-\left(x-\frac m2\right)^2+\frac{m^2}{4}+3.5$:

$$\frac m2=-1.5\rightarrow m=-3;$$

$$\frac{m^2}{4}+3.5=p$$

$$\rightarrow p=\frac{23}{4}$$

Answer: $m=-3,\quad p=\dfrac{23}{4}$

(a)

$$\begin{aligned}9x^2+6x-8&=0\\(3x-2)(3x+4)&=0\\x=\frac23\quad\text{or}\quad x&=-\frac43\end{aligned}$$

(b)

$$\begin{aligned}9(y-1)^2+6y-14&=0\\9(y-1)^2+6(y-1)+6-14&=0\\9(y-1)^2+6(y-1)-8&=0\end{aligned}$$

From (a),

$$\begin{aligned}y-1=\frac23\quad\text{or}\quad y-1&=-\frac43\\y=1\tfrac23\quad\text{or}\quad y&=-\tfrac13\end{aligned}$$

Answer: (a) $x=\dfrac23$ or $x=-\dfrac43$ (b) $y=1\frac23$ or $y=-\frac13$

(a)

$$\begin{aligned}&(x-3)(2x+8)=-12\\[0pt]&2x^2+8x-6x-24+12=0\\[0pt]&2x^2+2x-12=0\\[0pt]&x^2+x-6=0\\[0pt]&(x+3)(x-2)=0\\[0pt]&x=-3\ \ \textit{or}\ x=2\end{aligned}$$

(b)

$$\begin{aligned}&\frac4{2x-3}-\frac3{x+2}=1\\[9pt]&\frac{4(x+2)-3(2x-3)}{(2x-3)(x+2)}=1\\[9pt]&4(x+2)-3(2x-3)=(2x-3)(x+2)\\[9pt]&4x+8-6x+9=2x^2+x-6\\[9pt]&2x^2+3x-23=0\\[9pt]&x=\frac{-3\pm\sqrt{3^2-4(2)(-23)}}{2(2)}\\[9pt]&x=\frac{-3\pm\sqrt{193}}{4}\\[9pt]&x=\ 2.72\ \textit{or}\ -4.22\end{aligned}$$

Answer: (a) $x=-3$ or $x=2$ (b) $x=2.72$ or $x=-4.22$

(a)

Time taken to travel from A to B $=\dfrac{12}{x-3}$ h

Time taken to travel from B to C $=\dfrac9{x+3}$ h

$$\begin{aligned}\frac{12}{x-3}+\frac9{x+3}&=1\frac14\\\frac{12(x+3)+9(x-3)}{(x-3)(x+3)}&=\frac54\\48x+144+36x-108&=5(x^2-9)\\84x+36&=5x^2-45\\5x^2-84x-81&=0\quad\text{(shown)}\end{aligned}$$

(b)

$$x=\frac{84\pm\sqrt{(-84)^2-4(5)(-81)}}{2(5)}$$

$$x=\frac{84\pm\sqrt{8676}}{10}$$

$$x=17.7145\quad\text{or}\quad-0.91450$$

$$x=17.71\quad\text{or}\quad-0.91$$

(c)

The solution $x=-0.91$ must be rejected, as it results in the speed $x$ km/h, to be a negative value.

(d)

$$\begin{aligned}\text{Difference}&=\left(\frac{12}{17.7145-3}-\frac9{17.7145+3}\right)\text{ h}\\&=0.381044\text{ h}\\&=22.8626\text{ min}\\&=22\text{ min }52\text{ sec}\end{aligned}$$

Answer: (b) $17.71$ or $-0.91$   (c) $x=-0.91$ is rejected because a speed cannot be negative   (d) 22 minutes 52 seconds

(a)(i)

$\dfrac1x$

(a)(ii)

$\dfrac1{x-4}$

(b)

$$\begin{aligned}\frac6x+\frac6{x-4}&=1\\\frac{2x-4}{x(x-4)}&=\frac16\\x^2-4x&=12x-24\\x^2-16x+24&=0\end{aligned}$$

(c)(i)

$$\begin{aligned}x&=\frac{16\pm\sqrt{16^2-4(24)}}2\\&=\frac{16\pm\sqrt{160}}2\end{aligned}$$

$x=14.32$ or $x=1.68$

(c)(ii)

$x=1.68$ will make time taken to be negative for Tap B

(d)

$$\begin{aligned}14.32-4&=10.324h\\&=10\text{h}19\text{ mins}\end{aligned}$$

Answer: (a)(i) $\dfrac1x$   (ii) $\dfrac1{x-4}$   (c)(i) $x=14.32$ or $x=1.68$   (ii) $x=1.68$ is rejected because Tap B would then take a negative time   (d) 10 h 19 min