Quadratic Equations in One Unknown-Examples


  • Examples
    (a) Solve the equation \(x = \frac{{{x^2} + 37}}{{12}}\).
    (b) Hence solve the equation \(x = \frac{{{{(x + 1)}^2} + 25}}{{12}}\).
    (Express the answers as surds in their simplest form or in the form of \(a + bi\) if necessary, where a and b are real numbers.)
  • Solutions
    (a)
    \(\begin{array}{1}x = \frac{{{x^2} + 37}}
    {{12}}\\12x = {x^2} + 37\\{x^2} - 12x + 37 = 0\end{array}\)
    \(\begin{array}{1}x = \frac{{ - {\rm{ }}( - 12) \pm \sqrt {{{( - 12)}^2} - 4(1)(37)} }}{{2(1)}}\\
    = \frac{{12 \pm \sqrt { - {\rm{ }}4} }}{2}\\ = \frac{{12 \pm \sqrt 4 i}}{2}\\ = \underline{\underline {6 \pm i}} \end{array}\)

    (b)
    \(\begin{array}{1}x = \frac{{{{(x + 1)}^2} + 25}}{{12}}\\12x = {(x + 1)^2} + 25\\{(x + 1)^2} - 12x + 25 = 0\\{(x + 1)^2} - 12(x + 1) + 37 = 0\end{array}\)
    From the result of (a),
    \(\begin{array}{1}x + 1 = 6 \pm i\\
    x = \underline{\underline {5 \pm i}} \end{array}\)


  • Examples
    Mr. Wong deposits $40 000 in a bank, he gets the amount of $42 500 after two years. If the interest is compounded yearly, what is the interest rate per annum?
    (Give your answer correct to 3 significant figures.)
  • Solutions
    Let the interest rate per annum be r %.
    \(\begin{array}{1}40{\rm{ }}000{(1 + r\% )^2} = 42{\rm{ }}500\\40{\rm{ }}000{(1 + \frac{r}{{100}})^2} = 42{\rm{ }}500\\40{\rm{ }}000{(\frac{{100 + r}}{{100}})^2} = 42{\rm{ }}500\\40{\rm{ }}000(\frac{{{r^2} + 200r + 10{\rm{ }}000}}{{10{\rm{ }}000}}) = 42{\rm{ }}500\\{r^2} + 200r - 625 = 0\\r = \frac{{ - {\rm{ }}200 \pm \sqrt {{{200}^2} - 4(1)( - 625)} }}{{2(1)}}\\ = \frac{{ - {\rm{ }}200 \pm \sqrt {42{\rm{ }}500} }}{2}\end{array}\)
    \( = \underline{\underline {3.08}} \) (corr. to 3 sig. fig.) or -203 (corr. to 3 sig. fig.) (rejected)
    ∴ The interest rate per annum is 3.08%.


  • Examples
    Let x be the smaller number of two consecutive numbers. Three times the sum of the squares of the two numbers is greater than the square of the sum of the two numbers by 266.
    (a) Prove that \({x^2} + x - 132 = 0\).
    (b) Hence find the sum of the two numbers.
  • Solutions
    (a) Given that x is the smaller number of the two consecutive numbers, then the other number is \(x + 1\). \(\begin{array}{1}3[{x^2} + {(x + 1)^2}] - {[(x + 1) + x]^2} = 266\\3[{x^2} + ({x^2} + 2x + 1)] - {(2x + 1)^2} = 266\\6{x^2} + 6x + 3 - 4{x^2} - 4x - 1 = 266\\2{x^2} + 2x - 264 = 0\\{x^2} + x - 132 = 0\end{array}\)

    (b) \(\begin{array}{1}{x^2} + x - 132 = 0\\(x + 12)(x - 11) = 0\end{array}\)
    \(x = - {\rm{ }}12\) or \(x = 11\)
    When \(x = - {\rm{ }}12\),
    sum of the two numbers \(\begin{array}{l} = - {\rm{ }}12 + ( - 11)\\ = \underline{\underline { - {\rm{ }}23}} \end{array}\)
    When \(x = 11\),
    sum of the two numbers \(\begin{array}{l} = 11 + 12\\ = \underline{\underline {23}} \end{array}\)