### Quadratic Equation in One Unknown - (1.5-1.7)

#### Understand the relations between the discriminant of a quadratic equation and the nature of its roots

• 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 $${\bf{\Delta }} = {{\bf{b}}^2} - 4{\bf{ac}}\;\left\{ {\begin{array}{*{20}{c}}{ > 0\;\;\;2\;real\;roots\;\;\;}\\{ = 0\;\;\;1\;real\;root\;\;\;\;}\\{ < 0\;no\;real\;roots\;}\end{array}} \right.$$

• 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.

• 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

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

• $${\rm{x}}\left( {{\rm{x}} - 3} \right) = 54$$
$${{\rm{x}}^2} - 3x - 54 = 0$$
$$\left( {{\rm{x}} - 9} \right)\left( {{\rm{x}} - 6} \right) = 0$$
$${\rm{x}} = 9{\rm{\;or\;}}6\left( {{\rm{rejected}}} \right)$$
$$\therefore {\rm{x}} = 9{\rm{\;cm}}$$

#### Understand the relations between the roots and coefficients and form quadratic equations using these relations

• 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
$${\rm{\alpha \beta }} = \frac{{\rm{c}}}{{\rm{a}}}$$

• 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.
∴ $$\begin{array}{l} - \frac{k}{8} = 6\\\quad k = \underline{\underline { - 48}} \end{array}$$
$$\begin{array}{c}8{x^2} + ( - 48)x - 56 = 0\\{x^2} - 6x - 7 = 0\\(x - 7)(x + 1) = 0\\x = \underline{\underline 7} \;\,\,{\rm{or}}\quad x = \underline{\underline { - 1}} \end{array}$$