### More about equations - (5.1-5.4)

#### Use the graphical method to solve simultaneous equations in two unknowns, one linear and one quadratic in the form $$y = ax^2 + bx + c$$

• $$y=ax^2+bx+c$$
$$Ax+By+C=0$$

(Integers only)

$$\begin {cases}y=x^2-5\\ x-y=0\end{cases}$$

$$\begin{cases} x=\frac{1+\sqrt{21}}{2} ,\qquad y=\frac{1+\sqrt{21}}{2} \\ x=\frac{1-\sqrt{21}}{2}, \qquad y=\frac{1-\sqrt{21}}{2} \end{cases}$$

#### Use the algebraic method to solve simultaneous equations in two unknowns, one linear and one quadratic

• Solve the following simultaneous equations.
$$\left\{ \begin{array}{l}{x^2} + xy - 6 = 0\\3x + y - 7 = 0\end{array} \right.$$

• $$\left\{ \begin{array}{l}{x^2} + xy - 6 = 0{\kern 1pt} \;..................\;(1)\\3x + y - 7 = 0\;....................\;(2)\end{array} \right.$$
From (2), $$y = 7 - 3x\;...........\;(3)$$

Substitute (3) into (1),
$$\begin{array}{c}{x^2} + x(7 - 3x) - 6 = 0\\{x^2} + 7x - 3{x^2} - 6 = 0\\ - {\rm{ }}2{x^2} + 7x - 6 = 0\\2{x^2} - 7x + 6 = 0\\(x - 2)(2x - 3) = 0\end{array}$$
$$x = 2$$ or $$x = \frac{3}{2}$$

Substitute $$x = 2$$ into (3), $$y = 7 - 3(2)$$
$$= 1$$
Substitute $$x = \frac{3}{2}$$ into (3), $$y = 7 - 3(\frac{3}{2})$$
$$= \frac{5}{2}$$

The solutions are $$x = 2$$, $$y = 1$$ and $$x = \frac{3}{2}$$, $$y = \frac{5}{2}$$.

#### Solve equations (including fractional equations, exponential equations, logarithmic equations and trigonometric equations) which can be transformed into quadratic equations

• Solve the following equations.
(a)$$\frac{2}{{x - 2}} + \frac{x}{{4(x - 2)}} = \frac{{x - 4}}{4}$$

(b)$${2^{2x}} - {2^{x + 2}} - 5 = 0$$

(c)$${({\log _5}x)^2} + {\log _5}{x^3} - 4 = 0$$

(d)$$3\tan \theta + 2\cos \theta = 0$$

• $$\begin{array}{c}a)\qquad\frac{2}{{x - 2}} + \frac{x}{{4(x - 2)}} = \frac{{x - 4}}{4}\\\frac{{4(2)}}{{4(x - 2)}} + \frac{x}{{4(x - 2)}} = {\frac{{x - 4}}{4}^{}}\\\frac{{8 + x}}{{4(x - 2)}} = {\frac{{x - 4}}{4}^{}}\\8 + x = (x - 4){(x - 2)^{}}\\8 + x = {x^2} - 6x + 8\\{x^2} - 7x = 0\\x(x - 7) = {0^{}}\end{array}$$
$$x = \underline{\underline 0}$$ or $$x = \underline{\underline 7}$$

$$\begin{array}{c}b)\qquad{2^{2x}} - {2^{x + 2}} - 5 = 0\\{({2^x})^2} - ({2^x})({2^2}) - 5 = 0\\{({2^x})^2} - 4({2^x}) - 5 = 0\\({2^x} - 5)({2^x} + 1) = 0\end{array}$$
$${2^x} = 5$$ or $${2^x} = - 1$$ (rejected)
$$\begin{array}{c}\log {2^x} = \log 5\\x\log 2 = \log 5\\x = \frac{{\log 5}}{{\log 2}}\end{array}$$
$$x= \underline{\underline {2.32}}$$ (corr. to 3 sig. fig.)

$$\begin{array}{c}c)\qquad{({\log _5}x)^2} + {\log _5}{x^3} - 4 = 0\\{({\log _5}x)^2} + 3{\log _5}x - 4 = 0\\{({\log _5}x)^2} + 3({\log _5}x) - 4 = 0\\({\log _5}x - 1)({\log _5}x + 4) = 0\end{array}$$
$${\log _5}x = 1$$ or $${\log _5}x = - {\rm{ }}4$$
$$x = {5^1}$$ or $$x = {5^{ - {\rm{ }}4}}$$
Therefore, $$x = 5$$ or $$\frac{1}{{625}}$$.

$$\begin{array}{c}d)\qquad3\tan \theta + 2\cos \theta = 0\\\frac{{3\sin \theta }}{{\cos \theta }} + 2\cos \theta = 0\\3\sin \theta + 2{\cos ^2}\theta = 0\\3\sin \theta + 2(1 - {\sin ^2}\theta ) = 0\\2{\sin ^2}\theta - 3\sin \theta - 2 = 0\\(2\sin \theta + 1)(\sin \theta - 2) = 0\end{array}$$
$$2\sin \theta + 1 = 0$$ or $$\sin \theta - 2 = 0$$
$$\sin \theta = - \frac{1}{2}$$ or $$\sin \theta = 2$$ (rejected)

$$\theta = 180^\circ + 30^\circ$$ or $$360^\circ - 30^\circ$$
$$\theta = 210^\circ$$ or $$330^\circ$$

#### Solve problems involving equations which can be transformed into quadratic equations

• Solve the following equations.
(a) $$x - 6\sqrt x - 16 = 0$$
(b) $$6x - 19\sqrt x + 10 = 0$$

• a) Let $$v = \sqrt x$$, then $${v^2} = x$$.
$$\begin{array}{c} x - 6\sqrt x - 16 = 0\\{v^2} - 6v - 16 = 0\\(v - 8)(v + 2) = 0\end{array}$$
$$v = 8$$ or $$v = - 2$$
Since $$v = \sqrt x$$,
$$\sqrt x = 8$$ or $$\sqrt x = - 2$$ (rejected)
$$x = \underline{\underline {64}}$$

b) Let $$v = \sqrt x$$, then $${v^2} = x$$.
$$\begin{array}{c}6x - 19\sqrt x + 10 = 0\\6{v^2} - 19v + 10 = 0\\(3v - 2)(2v - 5) = 0\end{array}$$
$$v = \frac{2}{3}$$ or $$v = \frac{5}{2}$$
Since $$v = \sqrt x$$,
$$\sqrt x = \frac{2}{3}$$ or $$\sqrt x = \frac{5}{2}$$
$$x = \underline{\underline {\frac{4}{9}}} {\rm{ }}$$ or $$x = \underline{\underline {\frac{{25}}{4}}}$$