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Formula
Solve the equation
Number of solution
Relationship between roots and coefficients
Graph
$y = \left ( 2 x + 1 \right ) \left ( 3 x - 2 \right )$
$y = 0$
$x$Intercept
$\left ( - \dfrac { 1 } { 2 } , 0 \right )$, $\left ( \dfrac { 2 } { 3 } , 0 \right )$
$y$Intercept
$\left ( 0 , - 2 \right )$
Minimum
$\left ( \dfrac { 1 } { 12 } , - \dfrac { 49 } { 24 } \right )$
Standard form
$y = 6 \left ( x - \dfrac { 1 } { 12 } \right ) ^ { 2 } - \dfrac { 49 } { 24 }$
$\left( 2x+1 \right) \left( 3x-2 \right) = 0$
$\begin{array} {l} x = - \dfrac { 1 } { 2 } \\ x = \dfrac { 2 } { 3 } \end{array}$
Find solution by method of factorization
$\left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = 0$
 Expand the expression 
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = 0$
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = 0$
$\left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = \color{#FF6800}{ 0 }$
 If the product of the factor is 0, at least one factor should be 0 
$\begin{array} {l} \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
 Solve the equation to find $x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \end{array}$
$\begin{array} {l} x = - \dfrac { 1 } { 2 } \\ x = \dfrac { 2 } { 3 } \end{array}$
Find out the solution of the equation of the multiplication form
$\left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = \color{#FF6800}{ 0 }$
 If the product of the factor is 0, at least one factor should be 0 
$\begin{array} {l} \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
 Solve the equation to find $x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \end{array}$
$\begin{array} {l} x = - \dfrac { 1 } { 2 } \\ x = \dfrac { 2 } { 3 } \end{array}$
Solve the fractional equation
$\left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = \color{#FF6800}{ 0 }$
 Find solution by method of factorization 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \end{array}$
 2 real roots 
Find the number of solutions
$\left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = 0$
 Organize the expression 
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = 0$
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 }$
 Determine the number of roots using discriminant, $D=b^{2}-4ac$ from quadratic equation, $ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right )$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 6 \times \left ( - 2 \right )$
 Remove negative signs because negative numbers raised to even powers are positive 
$D = 1 ^ { 2 } - 4 \times 6 \times \left ( - 2 \right )$
$D = \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } - 4 \times 6 \times \left ( - 2 \right )$
 Calculate power 
$D = \color{#FF6800}{ 1 } - 4 \times 6 \times \left ( - 2 \right )$
$D = 1 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right )$
 Multiply the numbers 
$D = 1 + \color{#FF6800}{ 48 }$
$D = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 48 }$
 Add $1$ and $48$
$D = \color{#FF6800}{ 49 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 49 }$
 Since $D>0$ , the number of real root of the following quadratic equation is 2 
 2 real roots 
$\alpha + \beta = \dfrac { 1 } { 6 } , \alpha \beta = - \dfrac { 1 } { 3 }$
Find the sum and product of the two roots of the quadratic equation
$\left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = 0$
 Organize the expression 
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = 0$
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 }$
 In the quadratic equation $ax^{2}+bx+c=0$ , if the two roots are $\alpha, \beta$ , then it is $\alpha + \beta =-\dfrac{b}{a}$ , $\alpha\times\beta=\dfrac{c}{a}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 1 } { 6 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 2 } { 6 } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 1 } { 6 } } , \alpha \beta = \dfrac { - 2 } { 6 }$
 Solve the sign of a fraction with a negative sign 
$\alpha + \beta = \color{#FF6800}{ \dfrac { 1 } { 6 } } , \alpha \beta = \dfrac { - 2 } { 6 }$
$\alpha + \beta = \dfrac { 1 } { 6 } , \alpha \beta = \color{#FF6800}{ \dfrac { - 2 } { 6 } }$
 Reduce the fraction 
$\alpha + \beta = \dfrac { 1 } { 6 } , \alpha \beta = \color{#FF6800}{ \dfrac { - 1 } { 3 } }$
$\alpha + \beta = \dfrac { 1 } { 6 } , \alpha \beta = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } { 3 }$
 Move the minus sign to the front of the fraction 
$\alpha + \beta = \dfrac { 1 } { 6 } , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } }$
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