Quadratic Equation in One Unknown  (1.51.7)
Understand the relations between the discriminant of a quadratic equation and the nature of its roots

Theory
In the quadratic equation \({\bf{a}}{{\bf{x}}^2} + {\bf{bx}} + {\bf{c}} = 0\), the discriminate \({\bf{\Delta }}\) decides how many roots the equation has \({\bf{\Delta }} = {{\bf{b}}^2}  4{\bf{ac}}\;\left\{ {\begin{array}{*{20}{c}}{ > 0\;\;\;2\;real\;roots\;\;\;}\\{ = 0\;\;\;1\;real\;root\;\;\;\;}\\{ < 0\;no\;real\;roots\;}\end{array}} \right.\)

Examples
Let k be a constant. Express the discriminant of the quadratic equation \(5{\bf{k}}{{\bf{x}}^2} + 24{\bf{x}} + 6 = 0\) in terms of k.
If the equation has no real roots, find the range of values of k. 
Solutions
From the equation, \(\left\{ {\begin{array}{*{20}{c}}{a = 5k}\\{b = 24}\\{c = 6}\end{array}} \right.\)
\({\rm{\Delta }} = {{\rm{b}}^2}  4ac\)
\({\rm{\Delta }} = {\left( {24} \right)^2}  4\left( {5k} \right)\left( 6 \right)\)
\({\rm{\Delta }} = 576  120{\rm{k}}\)
the equation has no real root
\(\therefore {\rm{\Delta }} < 0\)
\({\rm{\Delta }} = {{\rm{b}}^2}  4ac < 0\)
\({\rm{k}} > \frac{{24}}{5}\)
solve problems involving quadratic equations

Examples
In the figure, the area of rectangle ABCD is \(54{\rm{ c}}{{\rm{m}}^2}\). Find the value of x.

Solutions
\({\rm{x}}\left( {{\rm{x}}  3} \right) = 54\)
\({{\rm{x}}^2}  3x  54 = 0\)
\(\left( {{\rm{x}}  9} \right)\left( {{\rm{x}}  6} \right) = 0\)
\({\rm{x}} = 9{\rm{\;or\;}}6\left( {{\rm{rejected}}} \right)\)
\(\therefore {\rm{x}} = 9{\rm{\;cm}}\)
Understand the relations between the roots and coefficients and form quadratic equations using these relations

Theory
If the root of the quadratic equation \({\rm{a}}{{\rm{x}}^2} + bx + c = 0\) are \({\rm{\alpha }}\) and \({\rm{\beta }}\),
then, the sum of roots
\({\rm{\alpha }} + {\rm{\beta }} =  \frac{{\rm{b}}}{{\rm{a}}}\)
and, the product of roots
\({\rm{\alpha \beta }} = \frac{{\rm{c}}}{{\rm{a}}}\) 
Examples
It is given that the sum of the roots of the quadratic equation \(8{{\rm{x}}^2} + {\rm{kx}}56 = 0{\rm{\;}}\)is 6.
(a) Find the value of k.
(b) Solve the quadratic equation. 
Solutions
(a) ∵ Sum of roots = 6
∴ \(\begin{array}{l}  \frac{k}{8} = 6\\\quad k = \underline{\underline {  48}} \end{array}\)
(b) By substituting k = –48 into the equation, we have
\(\begin{array}{c}8{x^2} + (  48)x  56 = 0\\{x^2}  6x  7 = 0\\(x  7)(x + 1) = 0\\x = \underline{\underline 7} \;\,\,{\rm{or}}\quad x = \underline{\underline {  1}} \end{array}\)