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Formula
Solve the equation
Number of solution
Relationship between roots and coefficients
Graph
$y = \left ( x + 3 \right ) \left ( 2 x - 1 \right )$
$y = 0$
$x$Intercept
$\left ( \dfrac { 1 } { 2 } , 0 \right )$, $\left ( - 3 , 0 \right )$
$y$Intercept
$\left ( 0 , - 3 \right )$
Minimum
$\left ( - \dfrac { 5 } { 4 } , - \dfrac { 49 } { 8 } \right )$
Standard form
$y = 2 \left ( x + \dfrac { 5 } { 4 } \right ) ^ { 2 } - \dfrac { 49 } { 8 }$
$\left( x+3 \right) \left( 2x-1 \right) = 0$
$\begin{array} {l} x = - 3 \\ x = \dfrac { 1 } { 2 } \end{array}$
Find solution by method of factorization
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = 0$
 Expand the expression 
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = 0$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = 0$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 0 }$
 If the product of the factor is 0, at least one factor should be 0 
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \end{array}$
 Solve the equation to find $x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 2 } } \end{array}$
$\begin{array} {l} x = - 3 \\ x = \dfrac { 1 } { 2 } \end{array}$
Find out the solution of the equation of the multiplication form
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 0 }$
 If the product of the factor is 0, at least one factor should be 0 
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \end{array}$
 Solve the equation to find $x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 2 } } \end{array}$
 2 real roots 
Find the number of solutions
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = 0$
 Organize the expression 
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = 0$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 }$
 Determine the number of roots using discriminant, $D=b^{2}-4ac$ from quadratic equation, $ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right )$
$D = \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } - 4 \times 2 \times \left ( - 3 \right )$
 Calculate power 
$D = \color{#FF6800}{ 25 } - 4 \times 2 \times \left ( - 3 \right )$
$D = 25 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right )$
 Multiply the numbers 
$D = 25 + \color{#FF6800}{ 24 }$
$D = \color{#FF6800}{ 25 } \color{#FF6800}{ + } \color{#FF6800}{ 24 }$
 Add $25$ and $24$
$D = \color{#FF6800}{ 49 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 49 }$
 Since $D>0$ , the number of real root of the following quadratic equation is 2 
 2 real roots 
$\alpha + \beta = - \dfrac { 5 } { 2 } , \alpha \beta = - \dfrac { 3 } { 2 }$
Find the sum and product of the two roots of the quadratic equation
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = 0$
 Organize the expression 
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = 0$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 }$
 In the quadratic equation $ax^{2}+bx+c=0$ , if the two roots are $\alpha, \beta$ , then it is $\alpha + \beta =-\dfrac{b}{a}$ , $\alpha\times\beta=\dfrac{c}{a}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 3 } { 2 } }$
$\alpha + \beta = - \dfrac { 5 } { 2 } , \alpha \beta = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 3 } } { 2 }$
 Move the minus sign to the front of the fraction 
$\alpha + \beta = - \dfrac { 5 } { 2 } , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } }$
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