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 \(y = ax^2 + bx + c\)

  • Theory
    \(y=ax^2+bx+c\)
    \(Ax+By+C=0\)

    (Integers only)

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

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

  • Examples
  • Solutions

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

  • Theory
  • Examples

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

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

  • Theory
  • Examples

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

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

  • Theory
  • Examples

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

  • Solutions
    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 \(x = \underline{\underline {\frac{{25}}{4}}} \)