Quadratic Equation in One Unknown - (1.1-1.4)

Solve quadratic equations by factor method

  • Examples

    Solve the following quadratic equations by factor method.
    \(64x^2 – 9 = 0\)

  • Solutions

    \(\begin{array}{c}64{x^2} - 9 = 0\\(8x + 3)(8x - 3) = 0\\8x + 3 = 0{\rm{ or 8}}x - 3 = 0\\x = \underline{\underline { - \frac{3}{8}}} {\rm{ }}{\kern 1pt} {\rm{or }}x = \underline{\underline {\frac{3}{8}}} \end{array}\)

Form quadratic equations from given roots (The given roots are confined to real numbers. )

  • Examples

    Given two roots of a quadratic equation are \(\frac{4}{5}\;\)and \(\frac{3}{4}\) respectively. Write the equation in the form \({\rm{a}}{{\rm{x}}^2} + bx + c = 0\)

  • Solutions

    \(\left( {{\rm{x}} - \frac{4}{5}} \right)\left( {x - \frac{3}{4}} \right) = 0\)
    \({{\rm{x}}^2} - \frac{4}{5}x - \frac{3}{4}x + \left( {\frac{4}{5} \cdot \frac{3}{4}} \right) = 0\)
    \({{\rm{x}}^2} - \frac{{31}}{{20}}x + \frac{3}{5} = 0\)
    \(20{{\rm{x}}^2} - 31x + 12 = 0\)

Solve the equation \({\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}} = 0\) by plotting the graph of the parabola \({\rm{y}} = {\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}\) and reading the x-intercepts.

  • Examples

    Given the graph of the quadratic function \({\rm{y}} = \frac{1}{4}{x^2} - \frac{9}{4}\)
    Solve \(\frac{1}{4}{x^2} - 4 = 0\)


  • Solutions


    From the given graph, drag the dotted line to y=0, it intersects at \(\left( {3,0} \right)\) and \(\left( { - 3,0} \right)\).
    \(\therefore \) The solution to \(\frac{1}{4}{x^2} - \frac{9}{4} = 0\) is 3 or -3.

solve quadratic equations by the quadratic formula

  • Theory

    If \({\bf{a}}{{\bf{x}}^2} + {\bf{bx}} + {\bf{c}} = 0,\;\;\)Then \({\bf{x}} = \frac{{ - {\bf{b}} \pm \sqrt {{{\bf{b}}^2} - 4{\bf{ac}}} }}{{2{\bf{a}}}}\)



  • Examples

    Solve \(\left( {3 - {\rm{x}}} \right)\left( {{\rm{x}} + 3} \right) = \frac{{\left( {{\rm{x}} + 7} \right)\left( {{\rm{x}} - 9} \right)}}{2}\)



  • Solutions

    \(\left( {3 - {\rm{x}}} \right)\left( {{\rm{x}} + 3} \right) = \frac{{\left( {{\rm{x}} + 7} \right)\left( {{\rm{x}} - 9} \right)}}{2}\)
    \(2\left( {9 - {{\rm{x}}^2}} \right) = {x^2} - 2x - 63\)
    \(18 - 2{{\rm{x}}^2} = {x^2} - 2x - 63\)
    \(3{{\rm{x}}^2} - 2x - 81 = 0\)
    \({\rm{x}} = \frac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4\left( 3 \right)\left( { - 81} \right)} }}{{2\left( { - 3} \right)}}\)
    \({\rm{x}} = \frac{{2 \pm \sqrt {976} }}{6}\)
    \({\rm{x}} = \frac{{1 \pm 2\sqrt {61} }}{3}\)